Science
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Aptitude
Test Problems
in Physics
Edited by
S.S. Krotov
QW
GB8
CBS PUBLISHERS & DISTRIBUTORS
4596/1 A, 11 Deuya Gan], New Delhi - 110 002 (INDIA)
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HBHQTQJIBCTBO <<Hay1<a» Momma
Aptitude
Test Problems
in Physics
Edited by
S.S. Krotov
W
@
CB8
CBS PUBLISHERS & DISTRIBUTORS
4596/1A, 11 Darya Gan], New Delhi- 110 002 (INDIA)
Contributing authors
A.I. Buzdin
V.A. Il’in
I.V. Krivchenkov
S.S. Krotov
N.A. Sveshnikov
Translated from Russian by
Natalia Wadhwa
First published 1990 ,
Revised from the 1988 Russian edition
Ha auenuticxox neuxe
Printed in the Union of Soviet Socialist Republics
© Hanarenacrao <<Hayna». Frtaanast pettammn
dmarmo-uareuaruqecnoi nnteparypu, 1988
© English translation, N. Wadhwa, 1990
CBS Pub. ISBN 81-239-0488.:6
Mir Pub. ISBN 5-03-001468-3
First Indian Reprint : 1996
This edition has been published in India by arrangement with
Mir Publishers, Moscow.
© English translation, N. Wadhwa, 1990
For sale in India only.
All rights reserved. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted in any form or by any
means, without the prior permission of the publisher.
Published by
Satish Kumar Jain for CBS Publishers & Distributors
4596/I-’*, ll Daryaganj, New Delhi-110002. (India)
Printed at
.l.S. Offset Printers, Delhi - I l0 05l
Contents
Preface 6
1. Mechanics 9
2. Heat and Molecular Physics 53
3. Electricity and Magnetism. 73
4. Optics 99
Solutions .......... 107
Preface
The present state of science and technology
is such that a large number of scientists
and engineers must be educated at an advanced level. This cannot be done without
significantly raising the level of teaching
physics, with an emphasis on the individual
and special efforts to detect and nurture
budding talents. In this respect, physics
olympiads for students at secondary school
and vocational training colleges are important in bringing to light the brightest students and in correctly guiding them in their
choice of profession.
This book, which is a collection of physics aptitude test problems, draws on the
experience of the physics olympiads conducted during the last fifteen years among
the schoolchildren of Moscow. A Moscow
physics olympiad includes three preliminary
theoretical rounds at the regional, city,
and qualifying levels, followed by a final
practical round. After the final round, a
team of Moscow schoolchildren is selected for
participation in the all-Union olympiad.
The complexity of the problems set for
each round increases gradually, starting
from the simplest problems at regional level, problems which can be solved simply
by havinga thorough knowledge of the basic laws and concepts of physics. The problems at the qualifying stage are much more
complicated. Some of the problems at this
level involve a certain amount of research
Preface 7
(as a rule, the problems make participation
in the olympiads even more challenging).
This collection contains problems from
the theoretical rounds only. The structure of
the book reflects the emphasis given to it
in different sections of the physics course at
such competitions. The number of problems
set at an olympiad and the fraction allotted
to a particular topic in the book are determined by the number of years the topic is
taught at school. A detailed classification
of different topics is not given since some
are represented by only one or two proble s,
while other topics have dozens of problems.
Most of the problems are originalg and a
considerable proportion of them w s.composed by the authors. The most difgcult
problems are marked by asterisks. feing
the product of a close group of authors, the
book reflects certain traditions and experience drawn from Moscow olympiads only.
A feature of the book is that it presents tlze
scientific views and working style of a
group of like-minded scientists.
In view of all this, the book should attract
a large circle of readers. The best way to
use it is as a supplementary material to the
existing collections of problems in elementary physics. It will be especially useful
to those who have gone through the general
physics course, and want to improve their
knowledge, or try their strength at nonstandard problems, or to develop an intuitive
approach to physics. Although it is recommended primarily for high-school_students,
we believe that college students in junior
classes will also lind something interesting
in it. The book will also be useful for organizers of physics study circles, lecturers
taking evening and correspondence courses,
and for teachers conducting extracurricular
activities.
This book would have never been put together without the inspiration of Academician I.K. Kikoin, who encouraged the
compilation of such a collection of problems.
For many years, Academici-an Kikoin
headed the central organizing committee
for the all-Union olympiads for schoolchildren and chaired the editorial board of the
journal Kvanst (Quant) and the series "Little Quant Library” The book is a mark of
our respect and a tribute to the memory of
this renowned Soviet scientist.
The authors would like to place on record
their gratitude to their senior colleagues in
the olympiad movement. Thanks are due to
V.K. Peterson, G.E. Pustovalov, G.Ya. Myakishev, A.V. Tkachuk, V.I. Grigor’ev,
and B.B. Bukhovtsev, who helped us in
the formation of our concepts about the
physical problem. We are also indebted to
the members of the jury of recent Moscow
olympiads, who suggested a number of the
problems included in this book. Finally, it
gives us great pleasure to express our gratitude to G.V. Meledin, who read through
the manuscript and made a number of helpful remarks and suggestions for improving
both the content and style of the book. _
Problems
1. Mechanics
For the problems of this chapter, the free-fall acceleration g (wherever required) should be taken
equal to 10 m/s“.
1.1. A body with zero initial velocity moves
down an inclined plane from a height h
and then ascends along the same plane
with an initial velocity such that it stops
at the same height h. In which case is the
time of motion longer?
1.2. At a distance L = 400 m from the
traffic light, brakes are applied to a locomotive moving at a velocity v = 54 km/h.
Determine the position of the locomotive
relative to the traffic light 1 min after the
application of brakes if its acceleration
a = -0.3 m/sz.
1.3. A helicopter takes off along the vertical with an acceleration a = 3 m/s2 and
zero initial velocity. In a certain time tl,
the pilot switches off the engine. At the
point of take-off, the sound dies away in a
Determine the velocity v of the helicopter at the moment when its engine is
10 Aptitude Test Problems in Physics
switched off, assuming that the velocity c
of sound is 320 m/s.
1.4. A point mass starts moving in a
straight line with a constant acceleration a.
At a time tl after the beginning of motion,
the acceleration changes sign, remaining
the same in magnitude.
Determine the time t from the beginning
of motion in which the point mass returns
to the initial position.
1.5. Two bodies move in a straight line towards each other at initial velocities v,
and U2 and with constant accelerations
al and aa directed against the corresponding
velocities at the initial instant.
What must be the maximum initial separation lmax between the bodies for which
they meet during the motion?
1.6. Two steel balls fall freely on an elastic
slab. The first ball is dropped from a height
h, = 44 cm and the second from a height
hz = 11 cm t s after the first ball. After
the passage of time 1*, the velocities of the
balls coincide in magnitude and direction.
Determine the time t and the time interval during which the velocities of the two
balls will be equal, assuming that the balls
do not collide.
1.7*. Small balls with zero initial velocity
fall from a height H = R/8 near the vertical axis of symmetry on a concave spherical surface of radius R.
Assuming that the impacts of the balls
against the surface are perfectly elastic,
prove that after the first impact each ball
1. Mechanics 14
gets into the lowest point of the spherical
surface (the balls do not collide).
1.8. A small ball thrown at an initial velocity vo at an angle oc to the horizontal strikes
a vertical wall moving towards it at
a horizontal velocity v and is bounced to
the point from which it was thrown.
Determine the time t from the beginning
of motion to the moment of impact, neglecting friction losses.
1.9*. A small ball moves at a constant velocity v along a horizontal surface and at
point A falls into a vertical well of depth H
and radius r. The velocity uof the ball forms
an angle ot with the diameter of the well
drawn through point A (Fig. 1, top view).
Determine the relation between v, H, r,
and ot for which the ball can "get out" of
the well after elastic impacts with the walls.
Friction losses should be neglected.
1.10. A cannon fires from under a shelter
inclined at an angle cx to the horizontal
(Fig. 2). The cannon is at point A at a distance Z from the base of the shelter (point
B). The initial velocity of the shell is vo,
and its trajectory lies in the plane of the
figure.
Determine the maximum range Lmax
of the shell.
1.11. The slopes of the windscreen of two
motorcars are fl, = 30° and [52 = 15° respectively.
· At what ratio vl/v, of the velocities of the
cars will their drivers see the hailstones
bounced by the windscreen of their cars in
12 Aptitude Test Problems in Physics
0. •
A °°. --...."A
Fig. 1
/’
.4%.
zA./·/
£L..é
Fig. 2
\\
§
&€g
Fig. 3
____ A
1/ P \\ V
T" '<¤
6/// ,.//, /.2. »/ » / x
Fig. 4
1. Mechanics 13
the vertical direction? Assume that hailstones fall vertically.
1.12. A sheet of plywood moves over _a
smooth horizontal surface. The velocities of
points A and B are equal to v and lie in the
plane of the sheet (Fig. 3).
Determine the velocity of point C.
1.13. A car must be parked in a small gap
between the cars parked in a row along the
pavement.
Should the car be driven out forwards or
backwards for the manoeuvre if only its
front wheels can be turned?
1.14*. An aeroplane flying along the horizontal at a velocity vo starts to ascend, describing a circle in the vertical plane. The
velocity of the plane changes with height h
above the initial level of motion according
to the law v2 = vf} -- 2a0h. The velocity of
the plane at the upper point of the trajectory is U1 = vo/2.
Determine the acceleration a of the plane
at the moment when its velocity is directed
vertically upwards.
1.15. An open merry—go—round rotates at an
angular velocity co. A person stands in it at
a distance r from the rotational axis. It
is raining, and the raindrops fall vertically
at a velocity vo.
How should the person hold an umbrella
to protect himself from the rain in the best
way?
1.16*. A bobbin rolls without slipping over
a horizontal surface so that the velocity v
of the end of the thread (point A) is directed
14 Aptitude Test Problems in *Physics
along the horizontal. A board hinged at
point B leans against the bobbin (Fig. 4).
The inner and oute1· radii of the bobbin are
r and R respectively.
Determine the angular velocity to of the
hoard as a function of an angle ot.
1.17. A magnetic tape is wound on an empty spool rotating at a constant angular velocity. The final radius rt of the winding was
found to be three times as large as the initial radius ri (Fig. 5). The winding time of
the tape is tl.
What is the time t, required for winding ·a
tape whose thickness is half that of the initial tape?
1.18. It was found that the winding radius
oi~ a tape on a cassette was reduced by half
in a time tl = 20 min of operation.
In what time tz will the winding radius
be reduced by half again?
1.19. Two 1·ings O and O' are put on two
vertical stationary rods AB and A’B'
respectively. An inextensible thread is fixed
at point A' and on ring O and is ·passed
through ring O' (Fig. 6).
Assuming that ring O' moves downwards
at a constant velocity vl, determine the velocity U2 of ring 0 if LAOO’ = cz.
1.20. A weightless inextensible rope rests
on a stationary wedge forming an angle cz
with the horizontal (Fig. 7). One end of the
rope is fixed to the wall at point A. A small
load is attached to the rope at point B.
The wedge starts moving to the right with
a constant acceleration 0,.
i. Mechanics 45
// ’_/ /
A A,
. X V"
0
r' 0 a'
Fig. 5 Fig. 6
A
i>
B
@
.%
\
\ x \\ ` \
Fig. 7
4*/- N"- im,.
1 .»P
i = @°
\ //
Fig. 8 Fig. 9
16 Aptitude Test Problems in Physics
Determine the acceleration al of the load
when it is still on the wedge.
1.21. An ant runs from an ant-hill in a
straight line so that its velocity is inversely
proportional to the distance from the centre
of the ant-hill. When the ant is at point A
at a distance ll = 1 m from the centre of
the ant-hill, its velocity vi = 2 cm/s.
What time will it take the ant to run
from point A to point B which is at a distance Z2 = 2 m from the centre of the anthill?
1.22. During the motion of a locomotive
in a circular path of radius R , wind is blowing in the horizontal direction. The trace
left by the smoke is shown in Fig. 8 (top
view).
Using the figure, determine the velocity
vwmd of the wind if it is known to be constant, and if the velocity vm of the locomotive is 36 km/h.
1.23*. Three schoolboys, Sam, John, and
Nick, are on merry-go-round. Sam and John
occupy diametrically opposite points on a
merry-go-round of radius r. Nick is on another merry-go-round of radius R. The positions of the boys at the initial instant are
shown in Fig. 9.
Considering that the merry-go-round
touch each other and rotate in the same direction at the same angular velocity oi,
determine the nature of motion of Nick
from J ohn’s point of view and of Sam from
Nick’s point of view.
1.24. A hoop of radius R rests on a horizon-
1. Mechanics 17
tal surface. A similar hoop moves past it
at a velocity v.
Determine the velocity vA of the upper
point of "intersection" of the hoops as a function of the distance d between their centres,
assuming that the hoops are_ thin, and the
second hoop is in contact with the first hoop
as it moves past the latter.
1.25. A hinged construction consists of
three rhombs with the ratio of sides 3:2:1
,· 7 6
25
A..U
.3
.5
6; 2
Fig. 10
(Fig. 10). Vertex A2 moves in the horizontal direction at a velocity v.
Determine the velocities of vertices A1,
A2, and B2 at the instant when the angles
of the construction are 90°
1.26. The free end of a thread wound on a
bobbin of inner radius r and outer radius R
is passed round a nail A hammered into
the wall (Fig. 11). The thread is pulled at a
constant velocity v.
Find the velocity v2 of the centre of the
bobbin at the instant when the thread forms
an angle oc with the vertical, assuming that
the bobbin rolls over the horizontal surface
without slipping.
2-0771
18 Aptitude Test Problems in Physics
1.27. A rigid ingot is pressed between two
parallel guides moving in horizontal directions at opposite velocities vi and v2.
At a certain instant of time, the points of
contact between the ingot and the guides
lie on a straight line perpendicular to the
directions of velocities v1 and U2 (Fig. 12).
A 1//
Al
a ·· Q
U2 I
wl {
Fig. 11 Fig, 12
What points ofthe ingot have velocities
equal in magnitude to U1 and U2 at this
instant?
1.28. A block lying on a long`f_ho1·izontal
conveyer belt moving at a constant velocity receives a velocity vo = 5 'm/s relative
to the-ground in the direction opposite to
the direction of motion of the conveyer. After t = 4 s, the velocity of the block becomes equal to the velocity of the belt. The
coefficient of friction between the block and
the belt is pi == 0.2.
Determine the velocity v of the conveyer
belt.
1.29. A body with zero initial velocity
slips from the top of an inclined plane
forming an angle ot with the horizontal. The
coefficient of friction it between the body
1, Mechanics 19
and the plane increases with the distance l
from the top according to the law it = bl.
The body stops before it reaches the end of
the plane.
Determine the time t from the beginning
of motion of the body to the moment when
it comes to rest.
1.30. A loaded sledge moving over ice gets
into a region covered with sand and comes
to rest before it passes half its length without turning. Then it acquires an initial
velocity by a jerk.
Determine the ratio of the braking lengths
and braking times before the first stop and
after the jerk.
1.31. A rope is passed round a stationary
horizontal log iixed at a certain height above
the ground. In order to keep a load of
mass m = 6 kg suspended on one end of the
rope, the maximum force T1 = 40 N
should be applied to the other end of the
rope.
Determine the minimum force T 2 which
must be applied to the rope to lift the load.
1.32. Why is it more difficult to turn the
steering wheel of a stationary motorcar
than of a moving car?
1.33. A certain constant force starts acting
on a body moving at a constant velocity
v. After a time inter al At, the velocity of
the body is reduced l§\Qhalf, and after the
same time interval, th— velocity is again
reduced by half.
Determine the velocity vt of the body after a time interval 3At from the moment
2*
20 Aptitude Test Problems in Physics
when the constant force starts acting.
1.34. A person carrying a spring balance
and a stopwatch is in a closed carriage standing on a horizontal segment of the railway.
When the carriage starts moving, the person sitting with his face in the direction of
motion (along the rails) and fixing a load of
mass m to the spring balance watches the
direction of the deflection of the load and
the readings of the balance, marking the
instants of time when the readings change
with the help of the stopwatch.
When the carriage starts moving and the
load is deileeted during the first time interval tl-.:4 sftowards the observer, the balance indicates a weight 1.25mg. During
the next time interval t2 = 3 s, the load
hangs in the vertical position, and
the balance indicates a weight mg. Then the
load is deflected to the left (across the carriage), and during an interval t3=25.12 s,
the balance again indicates a weight
1.25mg. Finally, during the last time interval i4 = 4 s, the load is deflected from the
observer, the reading of the balance
remaining the same.
Determine the position of the carriage relative to its initial position and its velocity by this instant of time, assuming that
the observer suppresses by his hand the oscillations resulting from a change in the
direction of deflection and in the readings
of the balance.
1.35. Two identical weightless rods are
hinged to each other and to a horizontal
1. Mechanics 21
beam (Fig. 13). The rigidity of each rod is
ko, and the angle between them is ,2oa.
Determine the rigidity k of the system of
rods relative to the vertical displacement
x ~// ··_. /<4Ag·j·gr//g zig; ·’¥’
kn fr.,
Z0:
A
Fig. 13
of a hinge A under the `action of a certain
force F, assuming that displacements are
small in comparison with the length of the
rods.
1.36. Two heavy balls are simultaneously
shot from two spring toy-guns arranged on a
horizontal plane at a distance s = 10 m
from each other. The first ball has the initial vertical velocity vi = 10 m/s, while
the second is shot at an angle cz to the horizontal at a velocity uz = 20 m/s. Each
ball experiences the action of the force of
gravity and the air drag F = uv, p. =
0.1 g/s.
Determine the angle on at which the balls
collide in air.
1.37. A light spring of length l and rigidity k is placed vertically on a table. A small
ball of mass m. falls on it.
Determine the height h from the surface
of the table at which the ball will have the
maximum velocity.
22 Aptitude Test Problems in Physics
1.38*. A heavy ball of mass m is tied to a
weightless thread of length l. The friction
of the ball against air is proportional to its
velocity relative to the air: F fr =- uv. A
strong horizontal wind is blowing at a constant velocity v.
Determine the period T of small oscillations, assuming that the oscillations of
the ball attenuate in a time much longer
than the period of oscillations.
1.39. A rubber string of mass m and rigidity lc is suspended at one end.
Determine the elongation Al of the string.
1.40. For the system at rest shown in
Fig. 14, determine the accelerations of all the
loads immediately after the lower thread
keeping the system in equilibrium has been
cut. Assume that the threads are weightless
and inextensible, the springs are weightless, the mass of the pulley is negligibly
small, and there is no friction at the point
of suspension.
1.41. A person hoists one of two loads of
equal mass at a constant velocity v (Fig. 15).
At the moment when the two loads are
at the same height h, the upper pulley is
released (is able to rotate without friction
like the lower pulley).
Indicate the load which touches the floor
first after a certain time t, assuming that the
person continues to slack the rope at the
same constant velocity v. The masses of
the pulleys and the ropes and the elongation of the ropes should be neglected.
1.42. A block can slide along an inclined
1. Mechanics 23
plane in various directions (Fig. 16). If
it receives a certain initial velocity v
directed downwards along the inclined
l
§§‘ x_\`\ I
._.. rhlr k ¤ lu
m2 @4 •
\ _\ \ Q\`\ \ \ `\~T{\\~.Q\\\
Fig. 14 Fig. 15
plane, its motion will be uniformly decelerated, and it comes to rest after traversing a
distance Z,. If the velocity of the same
Fig. 16
magnitude is imparted to it in the upward
direction. it comes to Trest after traversing
a. d`iS\:§£_I\9Q lg. At the. ·bo.ttom of the inclined
24 Aptitude Test Problems in Physics
plane, a perfectly smooth horizontal guide
is fixed.
Determine the distance l traversed by the
block over the inclined plane along the guide
if the initial velocity of the same magnitude is imparted to it in the horizontal direction?
1.43. A block is pushed upwards along the
roof forming an angle cx with the horizontal.
The time of the ascent of the block to the
upper point was found to be half the time
of its descent to the initial point.
Determine the coefficient of friction it
between the block and the roof.
1.44. Two balls are placed as shown in
Fig. 17 on a "weightless” support formed by
two smooth inclined planes each of which
forms an angle cx. with the horizontal. The
support can slide without friction along a
horizontal plane. The upper ball of mass
ml is released.
Determine the condition under which tl1e
lower ball of mass mz starts "climbing" up
the support.
1.45. A cylinder of mass m and radius r
rests on two supports of the same height
(Fig. 18). One support is stationary, while
the other slides from under the cylinder at
a velocity v.
Determ·ine the force of normal pressure
N sexerted by the cylinder on the stationary
support at the moment when the distance
between points A and B of the supports is
AB == r V2, assuming that the supports
werevery. close to each other at the initial
1. Mechanics 25
/777
MQA
Fig. 17
"`
.\ ¤>\ x\\ ~
Fig. 18
WW ln
py \\\
s
Fig. 19
instant. The friction between the cylinder
and the supports should be neglected.
1.46. A cylinder and a wedge with a vertical face, touching each other, move along
two smooth inclined planes forming the
same angle cz with the horizontal (Fig. 19).
The masses of the cylinder and the wedge
are ml and m, respectively.
Determine the force of normal pressure
N exerted by the wedge on the cylinder,
26 Aptitude Test Problems in Physics
neglecting the friction between them.
1.47. A weightless rod of length I with a
small load of mass m at the end is hinged at
point A (Fig. 20) and occupies a strictly vertical position, touching a body of mass•M.
A light jerk sets the system in motion.
’" i c
,,,,"/V
Fig. 20 s.
§
\ ~../»
s0
\.__/
rag. 21 A
For what mass ratio M/m will the rod
form an angle oc = at/6 with the horizontal
at the moment of the separation from the
body? What will be the velocity u of the
body at this moment? Friction should be
neglected.
1.48. A homogeneous rod AB of mass m
and length l leans with its lower end against
the wall and is kept in the inclined position
by a string DC (Fig. 21). The string is tied
at point C to the wall and at point D to the
rod so that AD = AB/3. The angles formed
by the string and the rod with the wall are
oc and [3 respectively.
1. Mechanics 27
Find all possible values of the coefficient
of friction it between the rod and the wall.
1.49*. A massive disc rotates about a verti.cal axis at an angular velocity Q. A smaller
disc of mass m and radius r, whose axis is
strictly vertical, is lowered on the first
disc (Fig. 22). The distance between the axes
of the discs is d (d> r), and the coefficient of friction between them is p.
Determine the steady-state angular velocity os of the smaller disc. What moment of
force 0/it must be applied to the axis of the
larger disc to maintain its velocity of rotation constant? The radius of the larger disc
is R > d + r. The friction at the axes of
the discs should be neglected.
u m'
/I/I 'Q cz //
8/ \g _ ar/2
;· sz m?
Fig. 22 Fig. 23
1.50. Two rigidly connected homogeneous
rods of the same length and mass ml and
mz respectively form an angle an/2 and rest
on a rough horizontal surface (Fig. 23).
The system is uniformly pulled with the
help of a string fixed to the vertex of the
angle and parallel to the surface.
28 Aptitude Test Problems in Physics
Determine the angle oc formed by the
string and the rod of mass ml.
1.51. A ball moving at a velocity v =
10 m/s hits the foot of a football player.
Determine the velocity u with which. the
foot should move for the ball impinging on
it to come to a halt, assuming that the
mass of the ball is much smaller than the
mass of the foot and that the impact is perfectly elastic.
1.52. A body of mass m freely falls to the
ground. A heavy bullet of mass M shot
along the horizontal hits the falling body
and sticks in it.
How will the time of fall of the body to
the ground change? Determine the time
tof fall if the bullet is known to hit the
body at the moment it traverses half the
distance, and the time of free fall from this
height is to. Assume that the mass of the
bullet is_ much larger than the mass of the
body (M > m). The air drag should be neglected.
Q--) U1
mi lh
.U2 élr
I _i
’”2
Fig. 24 Fig. 25
1.53. Two bodies of mass ml = 1 kg and
m, = 2 kg move towards each other in
mutually perpendicular directioxis at veloc-
1, Mechanics 29
ities U1 = 3 m/s and v, = 2 m/s (Fig. 24).
As a result of collision, the bodies stick together.
Determine the amount of heat Q liberated as a result of collision.
1.54. The inclined surfaces of two movable
wedges of the same mass M are smoothly
conjugated with the horizontal plane
(Fig. 25). A washer of massm slides down the
left wedge from a height h.
To what maximum height will the washer
rise along the right wedge? Friction should
be neglected.
1.55. A symmetric block of mass m,_ with
a notch of hemispherical shape of radius r
(
2
// ’, /. ,-’ " ’ M .-/*,-0, /
Fig. 26
rests on a smooth horizontal surface near the
wall (Fig. 26). A small washer of mass mz
slides without friction from the initial position.
Find the maximum velocity of the block.
1.56. A round box of inner diameterD containing a washer of radius r lies on a table
(Fig. 27). The box is 2w.·.ived as a whole at a
constant velocity U uirccéeel along the lines
30 Aptitude Test Problems in Physics
of centres of the box and the washer. At an
instant to, the washer hits the box.
Determine the time dependences of the
displacement xwash of the washer and of
its velocity vwash relative to the table,
starting from the instant to and assuming
:·-—-J-—--» we
lUI
P
_ \\\\\\\\\W\\&\
s. l s
··~·~···—···é········.:·
Fig. 27
that all the impacts of the washer against
the box are perfectly elastic. Plot the graphs
xwash (t) and vwash (t). The friction between the box and the washer should be
neglected.
1.57. A thin hoop of mass M and radius r
is placed on a horizontal plane. At the initial instant, the hoop is at rest. A small
washer of mass m with zero initial velocity
slides from the upper point of the hoop along
a smooth groove in the inner surface of the
hoop.
Determine the velocity u of the centre of
the hoop at the moment when the washer
is at a certain point A of the hoop, whose
radius vector forms an angle cp with the vertical (Fig. 28). The friction between the
hoop and the plane should be neglected.
1, Mechanics 31
1.58. A horizontal weightless rod of length
31 is suspended on two vertical strings. Two
loads of mass ml and mz are in equilibrium
at equal distances from each other and from
the ends of the strings (Fig. 29).
/ /’ /’ ’//’ _,, {1/.**,*/
Al
, , , , 1...4 4-.- se-* —»i
Fig. 28 Fig. 29
Determine the tension T of the left string
at the instant when the right string snaps.
1.59. A ring of mass m connecting freely
two identical thin hoops of mass M each
starts sliding down. The hoops move apart
over a rough horizontal surface.
Determine the acceleration a of the ring
at the initial instant if LAOIO2 = on
. · ,,L,.,_. .
F 02
A ll
Fig. 30
(Fig. 30), neglecting the friction between
the ring and the hoops.
32 Aptitude Test Problems in Physics
1.60. A flexible pipe of length l connects
two points A and B in space with an altitude difference h (Fig. 31). A rope passed
through the pipe is fixed at point A.
T1
*1
6O
Fig. 31
Determine the initial acceleration a of
the rope at the instant when it is released,
neglecting the friction between the rope
and the pipe walls.
1.61. A smooth washer impinges at a velocity v on a group of three smooth identical
blocks resting on a smooth horizontal _sur· *6)
Fig. 32
face as shown in Fig. 32. The mass of each
block is equal to the mass of the washer.
The diameter of the washer and its height
are equal to the edge of the block.
Determine the velocities of all the bodies
after the impact.
1. Mechanics 33
1.62. Several identical balls are at rest in
a smooth stationary horizontal circular
pipe. One of the balls explodes, disintegrating into two fragments of different masses.
Determine the final velocity of the body
formed as a result of all collisions, assuming that the collisions are perfectly
inelastic.
1.63. Three small bodies with the mass
ratio 3:4:5 (the mass of the lightest body
is m) are kept at three different points on
the inner surface of a smooth hemispherical
cup of radius r. The cup is fixed at its
lowest point on a horizontal surface. At a
certain instant, the bodies are released.
Determine the maximum amount of heat
Q that can be liberated in such a system.
At what initial arrangement of the bodies
will the amount of liberated heat be maximum? Assume that collisions are perfectly
inelastic.
1.64. Prove that the maximum velocity
imparted by an cx-particle to a proton
during their collision is 1.6 of the initial velocity of the oc-particle.
1.65. Why is it recommended that the air
pressure_in motorcar tyres be reduced. for
a motion of the motorcar over sand?
1.66. A long smooth cylindrical pipe of radius r is tilted at an angle oc to the horizontal (Fig. 33). A small body at point A is
pushed upwards along the inner surface of
the pipe so that the direction of its initial
velocity forms an angle cp with generatrix
AB.
:-0111
34 Aptitude Test Problems in Physics
Determine the minimum initial velocity
vo at which the body starts moving upwards
without being separated from the surface
of the pipe.
6
A ·5<> U°
-- E ......
Fig.33
1.67*. An inextensible rope tied to the axle
of a wheel of mass m and radius r is pulled
in the horizontal direction in the plane
of the wheel. The wheel rolls without jumping over a grid consisting of parallel horizontal rods arranged at a distance l from
one another (l <r).
Determine the average tension T of the
rope at which the wheel moves at a constant
velocity v, assuming the mass of the wheel
to be concentrated at its axle.
1.68. Two coupled wheels (i.e. light wheels
of radius r fixed to a thin heavy axle)
roll without slipping at a velocity v perpendicular to the boundary over a rough
horizontal plane changing into an inclined
plane of slope oc (Fig. 34).
Determine the value of v at which the
coupled wheels roll from the horizontal to
the inclined plane without being separated
from the surface.
1. Mechanics 35
1.69. A thin rim of mass m and radius r
rolls down an inclined plane of slope oz,
winding thereby a thin ribbon of li11ear den\ é·¤
(Z
Fig. 34 Fig. 35
sity p (Fig. 35). At the initial moment, the
rim is at a height h above the horizontal
surface. '
Determine the distance s from the foot
of the inclined plane at which the rim stops,
assuming that the inclined plane smoothly
changes into the horizontal plane.
1.70. Two small balls of the same size and
of mass ml and ml (ml > ml) are tied by a
thin weightless thread and dropped from a
balloon.
Determine the tension T of the thread during the flight after the motion of the balls
has become steady-state.
1.'¥ *. A ball is tied by a weightless inextensible thread to a fixed cylinder of 'radius r.
At the initial moment, the thread is wound
so that the ball touches the cylinder. Then
the ball acquires a velocity v in the radial
direction, and the thread starts unwinding
(Fig. 36). _
3i
36 Aptitude Test Problems in Physics
Determine; the length l of the unwound
segment of the thread by the instant of
time t, neglecting the force of gravity.
U
/*7 //1
{J »O
Fig. 36
1.72. Three small balls of the same mass,
white (w), green (g), and blue (b), are fixed
by weightless rods at the vertices of the
equilateral triangle with side l. The system
of balls is placed on a smooth horizontal surface and set in rotation about the centre of
mass with period T. At a certain instant,
the blue ball tears away from the frame.
Determine the distance L between the
blue and the green ball after the time T.
1.73. A block is connected to an identical
block through a weightless pulley by a
weightless inextensible thread of length 2l
(Fig. 37). The left block rests on a table at
a distance l from its edge, while the right
block is kept at the same level so that the
thread is unstretched and does not sag, and
then released.
What will happen first: will the left
block reach the edge of the table (and touch
the pulley) or the right block hit the table?
1. Mechanics 37
1.74. Two loads of the same mass are tied
to the ends of a weightless inextensible
J1
f
[il v
Fig. 37 Fig. 38
thread passed through a weightless pulley
(Fig. 38). Initially, the system is at rest,
and the loads are at the same level. Then
the right load abruptly acquires a horizontal velocity v in the plane of the figure.
Which load will be lower in a time?
1.75. Two balls of mass ml = 56 g and
mz = 28 g are suspended on two threads
of length ll = 7 cm and Z, -= 11 cm at the
end of a freely hanging rod (Fig. 39).
Determine the angular velocity co at which
the rod should be rotated about the vertical
axle so that it remains in the vertical position.
1.76. A weightless horizontal rigid rod along
which two balls of the same mass m can move
without friction rotates at a constant angular velocity co about a vertical axle. The
balls are connected by a weightless spring
of rigidity lc, whose length in the undeformed
state is lo. The ball which is closer to
the vertical axle is connected to it by the
same spring.
38 Aptitude Test Problems in Physics
Determine the lengths of the springs.
Under what conditions will the balls move
in circles?
EK
Vt- I
A ’”2
Fig. 39
1.77. Figure 40 shows the dependence of
the kinetic energy Wk of a body on the displacement s during the motion of the body
in a straight line. The force FA = 2 N
Wk
Y a ile PHP5;
60As
Fig. 40
-1. Mechanics 39
is known to act on the body at point A.
Determine the forces acting on the body
at point_s B and C.
|‘__.._....__{................
A
Fig. 41
1.78. A conveyer belt having a length l
and carrying a block of mass m moves at a
velocity v (Fig. 41).
Determine the velocity vo with which the
block should be pushed against the direction of motion of the conveyer so that the
amount of heat liberated as a result of deceleration of the block by the conveyer belt
is maximum. What is the maximum amount
of heat Q if the coefficient of friction is p
and the condition v < l/2ulg is satished?
1.79. A heavy pipe rolls from the same
height down two hills with different profiles
(Figs. 42 and 43). In the former case, the
Fig. 42 Fig. 43
40 Aptitude Test Problems in Physics
pipe rolls down without slipping, while in
the latter case, it slips on a certain region.
In what case will the velocity of the'pipe
at the end of the path be lower?
1.80. A heavy load is suspended on a light
spring. The spring is slowly pulled down at
the midpoint (a certain work A is done
thereby) and then released.
Determine the maximum kinetic energy
Wk of the load in the subsequent motion.
1.81. The masses of two stars are ml and
mz, and their separation is Z.
Determine the period T of their revolution in circular orbits about a common centre.
1.82. A meteorite approaching a planet
of mass M (in the straight line passing
through the centre of the planet) collides
with an automatic space station orbiting
the planet in the circular trajectory of radius R. The mass of the station is ten times
as large as the mass of the meteorite. As a
result of collision, the meteorite sticks in
the station which goes over to a new orbit
with the minimum distance R]2 from the
planet.
Determine the velocity u of the meteorite
before the collision.
1.83. The cosmonauts who landed at the
pole of a planet found that the force of
gravity there is 0.01 of that on the Earth,
while the duration of the day on' the planet
is the same as that on"the· Earth. It turned
out besides that the force of gravity on the
equator is zero.
1. Mechanics 41
Determine the radius R of the planet.
1.84. The radius of Neptune’s orbit is 30
times the radius of the Earth’s orbit.
Determine the period T N of revolution of
Neptune around the Sun.
1.85. Three loads of mass ml, mz, and M
are suspended on a string passed through
1 M mz
Fig. 44
two pulleys as shown in Fig. 44. The pulleys
are at the same distance from the points of
suspension.
Find the ratio of masses of the loads at
which the system is in equilibrium. Can
these conditions always be realized? The
friction should be neglected.
1.86. Determine the minimum coefficient
of friction umin between a thin homogeneous
rod and a floor at which a person can slowly
lift the rod from the floor without slippage
to the vertical position, applying to its
end a force perpendicular to it.
1.87. Three weightless rods of length I
each are hinged at points A and B lying on
the same horizontal and joint through hinges
at points C and D (Fig. 45). The length
42 Aptitude Test Problems in Physics
AB = 2l. A load of mass m, is suspended at
the hinge C.
Determine the minimum force F nlin applied to the hinge D for which the middle
rod remains horizontal.
gZ
0 g I}
gy
Fig. 45
1.88. A hexagonal pencil placed on an inclined plane with a slope oc at right angles
to the generatrix (i.e. the line of intersection of the plane and the horizontal surface)
remains at rest. The same pencil placed
parallel to the generatrix rolls down the
plane.
Determine the angle cp between the axis
of the pencil and the generatrix of the inclined plane (Fig. 46) at which the pencil
is still in equilibrium.
1.89. A homogeneous rod of length 2l
loans against a vertical wall at one end and
against a smooth stationary surface at another end.
What function y (x) must be used to describe the cross section of this surface for
the rod to remain in equilibrium in any
position even in the absence of friction?
Assume that the rod remains all the time
4, Mechanics 43
in the same vertical plane perpendicular
to the plane of the wall.
1.90. A thin perfectly rigid weightless rod
with a point-like ball fixed at one end is
deflected through a small angle on from its
/ l*{·`\___
. /6} `\
\ I \•
\ // :
\&....._ 1
Fig. 46 Fig. 47
equilibrium position and then released.
At the moment when the rod forms an
angle B < oc with the vertical, the ball undergoes a perfectly elastic collision with
an inclined wall (Fig. 47).
Determine the ratio T1/T of the period of
oscillations of this pendulum to tl1e period
of oscillations of a simple pendulum having
the same length.
1.91*. A ball of mass m falls from a certain
height on the pan of mass M (M > m) of a
spring balance. The rigidity of the sprin
is k.
Determine the displacement Ax of the
point about which the pointer of the balance will oseillate, assuming that the collisions of the ball with the pan are perfectly
elastic.
1.92. A bead of mass m can move without
friction along a long wire bent in a verti-
44 Aptitude Test Problems in Physics
cal plane in the shape of a graph of a certain function. Let lA be the length of the segment of the wire from the origin to a certain point A. It is known that if the bead
is released at point A such that ZA < lm,
its motion will be strictly harmonic:
Z (t) = IA cos cot.
Prove that there exists a point B(lA°<
l B) at which the condition of harmonicity
of oscillations will be violated.
1.93. Two blocks having mass m and 2m
and connected by a spring of rigidity lc
lie on a horizontal plane.
Determine the period T of small longitudinal oscillations of the system, neglecting
friction.
1.94. A heavy round log is suspended at
the ends on two ropes so that the distance
between the points of suspension of the
ropes is equal to the diameter of the log. The
length of each vertical segment of the ropes
is l.
Determine the period T of small oscillations of the system in a vertical plane perpendicular to the log.
1.95. Aload of mass M is on horizontal
rails. A pendulum made of a ball of mass m
tied to a weightless inextensible thread is
suspended to the load. The load can move
only along the rails.
Determine the ratio of the periods T2/T,
of small oscillations of the pendulum in vertical planes parallel and perpendicular to
the rails.
1.96. Four weightless rods of length l each
Mechanics 45
are connected by hinged joints and form a
,rhomb (Fig. 48). A hinge A is fixed, and a
load is suspended to a hinge C. Hinges D
and B are connected by a weightless spring
gf length 1.5l in the undeformed state. In
equilibrium, the rods form angles 010 = 30°
with the vertical.
Determine the period T of small oscillations of the load.
/’ / ,. / '///,
Ai
11
p5
11
1I/
Fig. 48 Fig. 49
L97. A thin hoop is hinged at point A
so that at the initial moment its centre of
mass is almost above point A (Fig. 49).
`Then the hoop is smoothly released, and in
a time 1: = 0.5 s, its centre of mass occupies the lowest position.
Determine the time t in which a pendulum formed by a heavy ball B fixed on a
weightless rigid rod whose length is equal
to the radius of the hoop will return to the
lowest equilibrium position if initially the
ball was near the extreme upper position
(Fig. 50) and was released without pushing.
46 Aptitude Test Problems in Physics
1.98. A weightless rigid rod with a load at
the end is hinged at point A to the wall so
that it can rotate in all directions (Fig. 51).
6
r• { 5
AI
LA
Fig. 50 Fig. 51
The rod is kept in the horizontal position
by a vertical inextensible thread of length
Z, fixed at its midpoint. The load receives a
momentum in the direction perpendicular
to the plane of the figure.
Determine the period T of small oscillations of the system.
1.99. One rope of a swing is fixed above the
other rope by b. The distance between the
poles of the swing is a. The lengths ll
and Z2 of the ropes are such that li —{— lg =
az + b2 (Fig. 52).
Determine the period T of small oscillations of the swing, neglecting the height of
the swinging person in comparison with
the above lengths.
1.100. Being a punctual man, the lift operator of a skyscraper hung an exact pendulum clock on the lift wall to know the end
of the working day. The lift moves with an
upward and downward accelerations during
the same time (according to a stationary
1. Mechanics 47
clock), the magnitudes of the accelerations
remaining unchanged.
Woill the operator finish his working day
in time, or will he work more (less) than
required?
1.101. The atmospheric pressure is known to
decrease with altitude. Therefore, at the up_._._...L_..-__..
0 ·Q>
ii
L1, _
» ,/IC//·..· A / .4.5-,/,’}‘,/*;,-95
Fig. 52
per storeys of the Moscow State University
building the atmospheric pressure must be
lower than at the lower storeys. In order to
verify this, a. student connected one arm of
a U—shaped manometer to the upper auditorium and the other arm to the lower auditorium.
What will the manometer indicate'?
1.102. Two thin-walled tubes closed at one
end are inserted one into the other and completely filled with mercury. The cross-sectional areas of the tubes are S and 2S.
48 Aptitude Test Problems in Physics
The atmospheric pressure is po == pmcrgh,
where omg,. is the density of mercury, g
is the free-fall acceleration, and h is the
height. The length of each tube is l> h.
What work A must be done by external
forces to slowly pull out the inner tube?
The pressure of mercury vapour and the
forces of adhesion between the material of the
tubes and mercury should be neglected.
1.103. Two cylinders with a horizontal
and a vertical axis respectively rest on a
horizontal surface. The cylinders are connected at the lower parts through a thin
tube. The "horizontal" cylinder of radius r
is open at one end and has a piston in it
Jé
" T- *
Z--; ii Qi
—;—; - - .—“—‘*--..-2
/ /»// , , / / / /
Fig. 53
(Fig. 53). The "vertical" cylinder is open at
the top. The cylinders contain water which
completely {ills the part of the horizontal
cylinder behind the piston and is at a certin level in the vertical cylinder.
Determine the level h of water in the vertical cylinder at which the piston is in equilibrium, neglecting friction.
1.104. An aluminium wire is wound on a
piece of cork of mass m,,0,k. The densities
pcmk, pal, and pw of cork, aluminium, and
4, Mechanics 49
water are 0.5 X 103 kg/m3, 2.7 >< 103 kg/m3,
and 1 >< 103 kg/m3 respectively.
Determine the minimum mass mal of
the wire that should be wound on the cork
so that the cork with the wire is completely
submerged in water.
1:105. One end of an iron chain is fixed to
a*‘sphere of mass M = 10 kg and of diameter D = 0.3 m (the volume of such a
sphere is V = 0.0141 m3), while the other
end is free. The length l of the chain is 3 m
and its mass m is 9 kg. The sphere with the
chain is in a reservoir whose depth H =
3° m.
Determine the depth at which the sphere
will float, assuming that iron is 7.85
times heavier than water.
1.106. Two bodies of the same volume but
of different masses are in equilibrium on a
lever.
Will the equilibrium be violated if the
lever is immersed in water so that the
bodies are completely submerged?
1.107. A flat wide and a high narrow box
float in two identical vessels filled with water. The boxes do not sink when two identical heavy bodies of mass m. each are
placed into them.
In which vessel will the level of water be
higher?
1.108. A steel ball floats in a vessel with
mercury.
How will the volume of the part of the
ball submerged in mercury change if a
4-0771
50 Aptitude Test Problems in Physics
layer of water completely covering the ball
is poured above the mercury?
1.109. A piece of ice floats in a vessel with
water above which a layer of a lighter oil
`is poured.
How will the level of the interface change
after the whole of ice melts? What will be
the change in the total level of liquid in the
vessel?
1.110. A homogeneous aluminium ball of
radius r ·= 0.5 cm is suspended on a weightless thread from an end of a homogeneous
rod of mass M = 4.4 g. The rod is placed
on the edge of a tumbler with water so that
half of the ball is submerged in water when
ar !
Fig. 54
the system is in equilibrium (Fig. 54). The
densities pa, and pw of aluminium and water
are 2.7 X 103 kg/ma and 1 >< 10?* kg/m°
respectively.
Determine the ratio y/x of the parts of
the rod to the brim, neglecting the surface tension on the boundaries between the
ball and water.
1.111. To what division will mercury fill
the tube of a freely falling barometer of
{1; Mechanics 51
length 105 cm at an atmospheric pressure
gf 760 mmHg?
1.112. A simple accelerometer (an instrument for measuring acceleration) can be made
in the form of a tube filled with a liquid
`F
1
"
,·r/4 rz/4
Fig. 55
and bent as shown in Fig. 55. During motion, the level of the liquid in the left arm
will be at a height hl, and in the right arm
at a height hz.
Determine the acceleration cz of a carriage
in which the accelerometer is installed,
assuming that the diameter of the tube is
much smaller than hl and hl.
1.113. A jet plane having a cabin of length
l` = 50 m flies along the horizontal with an
acceleration a =1 m/sz. The air density
in the cabin is p == 1.2 X 10*3 g/cm3.
What is the difference between the atmospheric pressure and the air pressure exerted on the ears of the passengers sitting in
the front, middle, and rear parts of the
cabin?
1.114. A tube filled with water and closed
at both ends uniformly rotates in a horizontal plane about the OO'-axis. The manometers fixed in the tube wall at distances rl
ga
52 Aptitude Test Problems in Physics
and r2 from the rotational axis indicate
pressures pl and pz respectively (Fig. 56),
O'!
\_l_,4w pr #02
0 """`”`"é"
Fig. 56
Determine the angular velocity to of rotation of the tube, assuming that the density pw of water is known.
1.115. Let us suppose that the dr.ag F to the
motion of a body in some medium depends
on the velocity v of the body as F = iw°°,
where cz > 0.
At what values of the exponent oz. will
the body pass an infinitely large distance
after an initial momentum has been imparted to it?
1.116. The atmospheric pressure on Mars is
known to be equal to 1/200 of the atmospheric pressure on the Earth. The diameter of
Mars is approximately equal to half the
Earth’s diameter, and the densities pE and
pM of the planets are 5.5 X 103 kg/ms
and 4 >< 103 kg/m3.
Determine the ratio of the masses of the
Martian and the Earth’s atmospheres.
—2. Heat and Molecular Physics
For the problems of this chapter, the universal
gas constant R (wherever required) should be
taken equal to 8.3 J/(mol·K).
2.1. Two vertical communicating cylinders
of different diameters contain a gas at a
constant temperature under pistons of mass
ml = 1 kg and mz = 2 kg. The cylinders
are in vacuum, and the pistons are at the
Smile height ho == 0.2 m.
What will be the difference .h in their
heights if the mass of the first piston is
made as large as the mass of the second
piston?
2.2. The temperature of the walls of a vessel containing a gas at a temperature T is
Twa1l·
In which case is the pressure exerted by
the gas on the vessel walls higher: when the
vessel walls are colder than the gas
(Twan < T) or when they are warmer than
th.B g°3S (TWa1l>
2.3. A cyclic process (cycle) 1-2-3-4-1
consisting of two isobars 2-3 and 4-1, isoclibr 1-2, and a certain process 3-4 representBd by a straight line on the p-V diagram
(Fig. 57) involves rz moles of an ideal gas.
The gas temperatures in states 1, 2, and
54 Aptitude Test Problems in Physics
3 are T1, T2, and T3 respectively, and points
2 and 4 lie on the same isotlierm.
Determine the work A done by the gas
during the cycle.
P ~`
\
2 `~
-- ss
T2
__ 4
T, `~`
`"`""`"""`“»7
Fig. 57
2.4. Three moles of- an ideal monatomic
gas perform. a cycle shown in Fig. 58. The
gas temperatures in different states are
·° 2 z
···~··~¢
Fig. 58
T3 = 400 K, T3 = 800 K, T3 = 2400 K,
and T4 = 1200 K.
Determine the work A done by the gas
during the cycle.
2.5. Determine the work A done by an ideal
gas during a closed cycle 1 -+4 -+3 -»
2. Heat and Molecular Physics 55
2 —>1 shown in Fig. 59 if pl = 10** Pa,
p0=3 X 105 Pa,p2=4'>< 10** Pa, V2V, = 10 1, and segments 4-3 and 2-1 of the
cycle are parallel to the V-axis.
P
e ---- ---1K*
pa ’-----"""'
P1 ···*· Y .2
' V, V2" V
Fig. 59
2.6. A gas takes part in two thermal p1·ocesses in which it is heated from the same initial state to the same final temperature.
P
`2
:5
\— 7
-—-——-—·-——-—··—-—··-—·*;;
Fig. 60
The processes are shown on the p-V diagram by straight lines 1-3 and 1-2
(Fig. 60).
Indicate the process in which the amount
of heat supplied to the gas is larger.
56 Aptitude Test Problems in Physics
2.7. A vessel of volume V = 30 l is separated into three `equal parts bystationary
semipermeable thin particles (Fig. 61). The
Fig. 61
left`, middle, and right parts are filled with
mH2 = 30 g of hydrogen, moz = 160 g of
oxygen, and mN2 = 70 g of nitrogen respectively. The left partition lets through only
hydrogen, while the right partition lets
through hydrogen and nitrogen.
What will be the pressure in each part of
the vessel after the equilibrium has been
set in if the vessel is kept at a constant
temperature T = 300 K?
2.8*. The descent module of a spacecraft
approaches the surface of a planet along the
vertical at a constant velocity, transmitting the data on outer pressure to the
spacecraft. The time dependence of pressure
(in arbitrary units) is shown in Fig. 62.
The data transmitted by the module after
landing are: the temperature T = 700 K
and the free-fall acceleration g = 10 m/s2.
Determine (a) the velocity v of landing of
the module if the atmosphere of the planet
is known to consist of carbon dioxide CO2,
and (b) the temperature Th at an altitude
h = 15 km above the surface of the planet.
I2, Heat and Molecular Physics 57
2.9. A vertical thermally insulated cylinder of volume Voontains n moles of an ideal
monatomic gas under a weightless piston.
p, arbitrary unils
00 A
.40
*20 2000 5000 as
Fig. 62A load of mass M is placed on the piston,
as a result of which the piston is displaced
by a distance h.
Determine the final temperature T, of
the gas established after the piston has
been displaced if the area of the piston is S
and the atmospheric pressure is po.
2.-10. A vertical cylinder of cross-sectional
area S contains one mole’of an ideal monatomic gas under a piston of mass M. At a
certain instant, a heater which transmits to
a· gas an amount of heat q per unit time is
switched on under the piston.
Determine the established velocity v
of the piston under the condition that the
gas pressure under the piston is constant
and equal to po, and the gas under the
piston is thermally insulated.
2.11*. The product of pressure and volume
(PV) of a gas does not change with volume
58 Aptitude Test Problems in Physics
at a constant temperature only provided that
the gas is ideal.
Will the product pV be higher or lower
under a very strong compression of a gas if
no assumption is made concerning the ideal
nature of the gas?
2.12*. A horizontal cylindrical vessel of
length 2l is separated by a thin heat-insulating piston into two equal parts each of
which contains n moles of an ideal monatomic gas at a temperature T. The piston is
connected to the end faces of the vessel by unI 21 E I
w
A5
Z
wtw
0
Fig. -63
deformed springs of rigidityk each (Fig. 63).
When an amount of heat Q is supplied
to the gas in the right part, the piston is
displaced to the left by a distance x = Z/2.
Determine the- amount of heat Q' given
away at the temperature T to a thermostat
with which the gas in the left part is in
thermal contact all the time.
2.13. A thermally insulated vessel is divided into two parts by a heat-insulating piston which can move in the vessel without
friction. The left part of the vessel contains
one mole of an ideal monatomic gas, and
2,. Heat and Molecular Physics 59
the right part is empty. The piston is connected to the right wall of the vessel
through a spring whose length in free state is
equal to the length of the vessel (Fig. 64).
Fig. 64
Determine the heat capacity GT of the system, neglecting the heat capacities of the
vessel, piston, and spring.
2.14. Prove that the efficiency of a heat engine based on a cycle consisting of two isotherms and two isochors is lower than the
efficiency of Ca1·not’s heat engine operating
with the same heater and cooler.
2.15*. Let us suppose that a planet of mass
M and radius r is surrounded by an atmosphere of constant density, consisting of
a gas of molar mass p.. _
Determine the temperature T of the atmosphere on the surface of the planet if the
height of the atmosphere is h (h < r).
2.16. It is known that the temperature in
the room is +20 °C when the outdoor temperature is -20 °C, and +10 °C when the
outdoor temperature is -40 °C.
Determine the temperature T of the radiator heating the room.
60 Aptitude Test Problems in Physics
2.17. A space object] has the shape of a
sphere of radius R. Heat sources ensuring
the heat evolution at a constant rate are
distributed uniformly over its volume. The
amount of heat liberated by a unit surface
area is proportional to the fourth power of
thermodynamic temperature.
In what proportion would the temperature of the object change if its radius decreased by half?
2.18*. A heat exchanger of length l consists of a tube of cross-sectional area 2S
with another tube of cross-sectional area
l Ti, l I 7`§, I
—->U
—->U I
L—»#-J
Fig. 65
S passing through it (Fig. 65). The walls of
the tubes are thin. The entire system is thermally insulated from the ambient. A liquid of density p and specific heat c is pumped
at a velocity v through the tubes. The
initial temperatures of the liquid in the
heat exchanger are Tu and T 12 respectively.
Determine the final temperatures T,1
and Tu of the liquid in the heat exchanger
2. Heat and Molecular Physics 61
if the liquid passes through the tubes in
the counterflow, assuming that the heat
transferred per unit time through a unit
area element is proportional to the temperature difference, the proportionality factor
being lc. The thermal conductivity of the
liquid in the direction of its flow should be
neglected.
2.19*. A closed cylindrical vessel of base
area S contains a substance in the gaseous
state outside the gravitational field of the
Earth. The mass of the gas is M and its
pressure is p such that p <pSat, where
peat is the saturated vapour pressure at a
given temperature. The vessel starts moving with an acceleration a directed along
the axis of the cylinder. The temperature is
maintained constant.
Determine the mass mliq of the liquid
condensed as a result of motion in the vessel.
2.20. The saturated water vapour pressure
on a planet is po = 760 mmHg.
p Determine the vapour density p.
2.21. In cold weather, water vapour can be
seen in the exhaled air. If the door of a warm
hut is opened on a chilly day, fog rushes into the hut.
Explain these phenomena.
2.22*. A vessel of volume V = 2 l contains
mH, = 2 g of hydrogen and some amount
of water. The pressure in the vessel is
pi = 17 X 105 Pa. The vessel is heated so
that the pressure in it increases to pr =
26 X 10** Pa, and wat_er partiallyevapo-
62 Aptitude Test Problems in Physics
rates. The molar mass of water vapour is
it = 18 X 10*3 kg/mol.
Determine the initial T, and final T,
temperature of water and its mass Am.
Hint. Make use of the following temperature dependence of the saturated water vapour pressure:
T, °C 100 120 133 152 180
psat, ><10¤ Pa 1 2 3 5 10
2.23. The lower end of a capillary of radius
r=.-.0.2 mm and length l=8cm is immersed
in water whose temperature is constant
and equal to Tm, = 0 °C. The temperature
of the upper end of the capillary is Tm, -—-=
100 °C.
Determine the height h to which the water in the capillary rises, assuming that
the thermal conductivity of the capillary
is much higher than the thermal conductivity of water in it. The heat exchange with
the ambient should be neglected.
Hint. Use the following temperature dependence of the surface tension of water:
T, °C 0 20 50 90
o, mN/m 76 73 67 60
2.24. A cylinder with a movable piston
contains air under a pressure pl and a soap
bubble of radius r. The surface tension is
o, and the temperature T is maintained constant.
Determine the pressure p2 to which the
air should be compressed by slowly pull-
5*2,. Heat and Molecular Physics 63
mg [the [piston into the cylinder for the soap
bubblefto reduce its size by half.
2.25. Why isi clay used instead of cement
(which has a higher strength) in laying
bricks for a hreplace? (Hint. Red-clay bricks
are used for building ii`r`eplaces..)
2.26. A thermally insulated vessel contains
two liquids with initial temperatures T1
and T2 and specific heats C1 and c2, separated by a nonconducting partition. The partition is removed, and the difference between the initial temperature of one of the
liquids and the temperature T established
in the vessel turns out to be equal to half
the difference between the initial temperatures of the liquids.
Determine the ratio ml/mz of the masses
of the liquids.
2.27. Water at 20 °C is poured into a test
tube whose bottom is immersed in a large
amount of water at 80 °C. As a result, the
water in the test tube is heated to 80 °C
during a time tl. Then water at 80 °C is
poured into the test tube whose bottom is
imm·ersed in a large amount of water at
20 °C. The water in the test tube is cooled to
20 °C during a time tz.
What time is longer: tl or tz?
2.28. The same mass of water is poured into
two identical light metal vessels. A heavy
ball (whose mass is equal to the mass of water and whose density is much higher than
that of water) is immersed on a thin nonconducting thread in one of the vessels so
that it is at the centre of the volume of the
64 Aptitude Test Problems in Physics
water in the vessel. The vessels are heated
to the boiling point of water and left to
cool. The time of cooling for the vessel
with the ball to the temperature of the
ambient is known to be lc times as long as
the time of cooling for the vessel without a
ball.
Determine the ratio cb/cw of the specific
heats of the ball material and water.
2.29. Two identical thermally insulated
cylindrical calorimeters of height h =
75 cm are filled to one-third. The first calorimeter is filled with ice formed as a result
of freezing water poured into it, and the
second is filled with water at Tw=10 °C.
Water from the second calorimeter is poured
into the first one, and as a result it becomes to be filled to two-thirds. After the
temperature has been stabilized in the first
calorimeter, its level of water increases by
Ah = 0.5 cm. The density of ice is pm, =0.9pw, the latent heat of fusion of ice is
lt = 340 kJ/kg, the specific heat of ice is
cm, = 2.1 kJ/(kg·K), and the specific
heat of water is cw =· 4.2 kJ/(kg·K).
Determine the initial temperature Tm
of ice in the first calorimeter.
2.30*. A mixture of equal masses of water
and ice (m = mw = mic, = 1 kg) is contained in a thermally insulated cylindrical
vessel under a light piston. The pressure on
the piston is slowly increased from the initial value po = 105 Pa to pl = 2.5 ><
10° Pa. The specific heats of water and
ice are cw = 4.2 kJ/(kg·K) and 0,,,,, =
2. Heat and Molecular Physics 65
2.1 kJ /(kg·K), the latent heat of fusion of
ice is 7t = 340 kJ/kg, and the density of-ice
is pic., = 0.9pw (where pw is the density
of water).
Determine the mass Am of ice which melts
in the process and the work A done by an
external force if it is known that the pressure required to decrease the fusion temperature of ice by 1°C is p == 14 X 10° Pa,
while the pressure required to reduce the
volume of a certain mass of water by 1%
is p' = 20 X 10** Pa.
(1) Solve the problem, assuming that water and ice are incompressible.
(2) Estimate the correction for the compressibility, assuming that the compressibility of ice is equal to half that value
for water.
2.31. It is well known that if an ordinary
water is salted, its boiling point rises.
Determine the change in the density of
saturated water vapour at the boiling
point.
2.32. For many substances, there exists a
temperature Tu and a pressure pt,. at which
all the three phases of a substance (gaseous,
liquid, and solid) are in equilibrium. These
temperature and pressure are known as the
triple point. For example, Tt, = 0.0075 °C
and pt,. = 4.58 mmHg for water. The ‘la·tent heat of vaporization of water at the
triple point is q = 2.48 X 103 kJ/kg, and
the latent heat of fusion of ice is A == 0.34 X
10“ kJ/kg.
Find the latent heat v of sublimation
6-0771
`66 Aptitude Test Problems in Physics
(i.e. a direct transition from the solid to
the gaseous state) of water at the triple
point.
2.33. The saturated vapour pressure above
an aqueous solution of sugar is known to be
lower than that aboye pure water, where
it is equal to put, by Ap = 0.05pS8tc,
Fig. 66
where c is the molar concentration of the
solution. A cylindrical vessel filled to
height hl--.:10 cm with a sugar solution of
concentration cl == 2 X 10* is placed under a wide bell. The same solution of concentration c, :10* is poured under the
hell to a level h, <h1 (Fig. 66).
Determine the level h of the solution in
the cylinder after the equilibrium has been
set in. The temperature is maintained constant and equal to 20 °C. The vapour above
the surface of the solution contains only
water molecules, and the molar mass of
water vapour is p. = 18 >< 10* kg/mol.
2.34. A long vertical brick duct is filled
with cast iron. The lower end of the duct
is maintained at a temperature T1 >
»z. Heat and Molecular Physics 67
Tmm (Tmm is the melting point of cast
iron), and the upper end at a temperature
fz < Tmclt. The thermal conductivity of
molten (liquid) cast iron is le times higher
than that of solid cast iron.
'T Determine the fraction of the duct filled
with molten metal.
2.35*. The shell of a space station is a
blackened sphere in which a temperature
T=-.500 K is maintained due to the operation of appliances of the station. The amount
of heat given away from a unit surface
area is proportional to the fourth power of
thermodynamic temperature. p
Determine the temperature Tx of the
shell. if the station is enveloped by a th.in
spherical black screen of nearly the same
radius as the radius of the shell.
2.36. A bucket contains a mixture of water
and ice of mass m = 10 kg. The bucket is
vc
ill
2-,
I
1I
I
0 20 40 00; min
Fig. 67
brought into a room, after which the temperature of the mixture is immediately
measured. The obtained T (1:) dependence
is plotted in Fig. 67. The specific heat of
68 Aptitude Test Problems in Physics
water is cw = 4.2 I/(kg·K), and the latent
heat of fusion of ice is it = 340 kJ/kg.
Determine the mass mm of ice in the
bucket at the moment it is brought in the
room, neglecting the heat capacity of the
bucket.
2.37*. The properties of a nonlinear resistor were investigated in a series of experiments. At first, the temperature dependence of the resistor was studied. As the
temperature was raised to T2 = 100 °C,
the resistance changed jumpwise from
R2 = 50 Q to R2 = 100 Q. The reverse abrupt change upon cooling took place at
T2 = 99 °C, Then a d.c. voltage U1 =
60 V was applied to the resistor. Its temperature was found to be T2 -··= 80 °C, Finally, when a d.c. voltage U2 =80V was
applied to the resistor, spontaneous current
oscillations were observed in the circuit.
The air temperature To in the laboratory
was constant and equal to 20 °C. The heat
transfer from the resistor was proportional
to the temperature difference between the
resistor and the ambient, the heat capacity
of the resistor being C = 3 J /K.
Determine the period T of current oscillations and the maximum and minimum
values of the current.
2.38. When raindrops fall on a red·brick
wall after dry and hot weather, hissing
sounds are produced.
Explain the phenomenon.
2.39. A thin U-tube -sealed at one end consists of three bends of length lt =-·· 250 mm
2. Heat and Molecular Physics 69
each, forming right angles. The vertical
parts of the tube are filled with mercury
to half the height (Fig. 68). All of mercury
~¤—°°#+:a
l
E
FL
l __ _L________ ___i
|" 1
Fig. 68
can be displaced from the tube by heating
slowly the gas in the sealed end of tl1e tube,
which is separated from the atmospheric
air by mercury.
Determine the work A done by the gas
thereby if the atmospheric pressure is po:
105 Pa, the density of mercury is pm, ==
13.6 >< 103 kg/m3, and the cross-sectional
area of the tube is S = 1 cm2.
2.40. The residual deformation of an elastic rod can be roughly described by using
the following model. If the elongation of
the rod Al < :::0 (where xo is the quantity
present for the given rod), the force required
to cause the elongation Al is determined
by Hooke’s law: F = k AZ, where k is the
rigidity of the rod. If Al > xm, the force does
not depend on elongation any longer (the
material of the rod starts "flowing"). If the
70 Aptitude Test Problems in Physics
load is then removed, the elongation of the
rod will decrease along CD which for the
sake of simplicity will be taken straight and
parallel to_ the segment AB (Fig. 69). Therer
00
{I
i/i7E
i:
'1
Ia
Ul I
II
II
II
. .-...l. ’
A xc I .1: Al
Fig. 69
fore, after the load has been removed completely, the rod remains deformed (point
D in the figure). Let us suppose that the rod
is initially stretched by Al == sc > xo and
then the load is removed.
Determine the maximum change AT in
the rod temperature if its heat capacity is
C, and the rod is thermally insulated.
2.41. A thin-walled cylinder of mass m,
height h, and cross—sectional area S is filled
with a gas and floats on the surface of water
(Fig. 70). As a result of leakage from the
lower part of the cylinder, the depth of its
submergence has increased by Ah.
Determine the initial pressure pl of the
gas in the cylinder if the atmospheric pressure is po, and the temperature remains
unchanged.
,,2. Heat and Molecular Physics 71
2.42. A shock wave is the region of an elevated pressure propagating in the positive
direction of the :1:-axis at a high velocity v.
:*3 •l·"m’S U ";
Fig. 70
p
pn -°°`-°_—_—`°°
Fig. 71
a
Fig. 72
At the moment of arrival of the wave, the
pI‘GSSl11‘G Bbfuptly iIlG1‘98S8S. This dBpBI1dence is plotted in Fig. 71.
72 Aptitude Test Problems in Physics
Determine the velocity u acquired by a
wedge immediately after the shock front
passes through it. The mass of the wedge is
m, and its size is shown in Fig. 72. Friction
should be neglected, and the velocity acquired by the wedge should be assumed to be
much lower than the velocity of the wave
(u < v).
$3, Electricity and Magnetism
For the problems of this chapter, assume (wherever
required) that the electric constant eo is specified.
3.1. A thin insulator rod is placed between
two unlike point charges —l—q1 and —q2
(Fig. 73).
® [Z;] G->
gz Q2
Fig. 73
How will the forces acting on the charges
change?
3.2. An electric field line emerges from a
positive point charge —}-q, at an angle on to the
A-.
ff!
[Ogg
9/
Fig. 74
straight line connecting it to a negative
point charge -—q2 (Fig. 74).
At what angle B will the field line enter
the charge —-q,?
74 Aptitude Test Problems in Physics
3.3. Determine the strength E of the electric
field at the centre of a hemisphere produced
by charges uniformly distributed with a
density o' over the surface of this hemisphere.
3.4. The strength of the electric field produced by charges uniformly distributed over
the surface of a hemisphere at its centre O
is E 0. A part of the surface is isolated from
Fig. 75
this hemisphere by two planes passing
through the same diameter and forming
an angle oc with each other (Fig. 75).
Determine the electric iield strength E
produced at the same point O by the charges
located on .the isolated surface (on the
"mericarp").
3.5*. Two parallel-plate capacitors are arranged perpendicular to the common axis.
The separation d between the capacitors is
much larger than the separation l between
their plates and than their size. The capacitors are charged to q1 and q2 respectively
(Fig. 76).
Find the force F of interaction between
the capacitors.
3.6. Determine the force F of interaction
between two hemispheres of radius R touch-
Electricity and Magnetism 75
*5:
mg teach other along the equator if one
,.hemisphere is uniformly charged with a
surface density U1 and the other with a surface density oz.
,____zz
`Fig. 76
3.7. The minimum strength of a uniform
electric field which can tear a conducting
uncharged thin-walled sphere into two parts
is known to be E0.
•Determine the minimum electric field
strength El required to tear the sphere of
twice as large radius if the thickness of its
walls is the same as in the former case.
3.8. Three small identical neutral metal
balls are at the vertices of an equilateral
triangle. The balls are in turn connected
to an isolated large conducting sphere whose
centre is on the perpendicular erected
from the plane of the triangle and passing
through its centre. As a result, the first and
second balls have acquired charges ql and
q, respectively.
Determine the charge qa of the third ball.
3.9. A metal sphere having a radius rl
charged to a potential cpl is enveloped by a
thin-walled conducting spherical shell of
radius r, (Fig. 77).
I6 Aptitude Test Problems in Physics
Determine the potential (pz acquired by
the sphere after it has been connected for a
short time to the shell by a conductor.
Fig. 77
3.10. A very small earthed conducting sphere
is at a distance a from a point charge cj,
and at a distance b from a point charge
q2 (a<b). At a certain instant, the sphere
starts expanding so that its radius grows
according to the law R s= vt.
Determine the time dependence I (t) of
the current in the earthing conductor, assuming that the point charges and the centre
of the sphere are at rest, and in due time
the initial point charges get into the expanding} sphere without touching it (through
small holes). '
3.11. Three; uncharged capacitors of capacitance C1, C2, and C3 are connected as
shown in Fig. 78 to one another and to
points A, B, and D at potentials cpA, cpB, and
(PD- `
Determine the potential (po at point O.
3. Electricity and Magnetism 77
3.12*. The thickness of a flat sheet of a metal foil; is d, and its area is S. A charge q
is located at a distance Z from the centre of
the sheet such that d < 1/S <l.
Determine the force F with which the
sheet is attracted to the charge q, assuming
that the straight line connecting the charge
to the centre of the sheet is perpendicular
to the surface of the sheet.
I
61 A
A .,6 l B P3 »
/"°* K [1--1
JU
Fig. 78 Fig. 79
3.13. Where must a current source be connected to the circuit shown in Fig. 79 in
order to charge all the six capacitors having
equal capacitances?
23.14. A parallel-plate capacitor is iilled by
.a dielectric whose permittivity varies with
ithe applied voltage according to the law
it = ctU, where cc = 1 V"1. The same (but
containing no dielectric) capacitor charged
to a voltage U0 = 156 V is connected in
parallel to the first "nonlinear" uncharged
capacitor.
Determine the final- voltage U across the
capacitors.
78 Aptitude Test Problems in Physics
3.15. Two small balls of mass m, bearing a
charge q each, are connected by a nonconducting thread of length 2l. At a certain instant, the middle of the thread starts moving at a constant velocity v perpendicular
to the direction of the thread at the initial
instant.
Determine the minimum distance d between the balls.
3.16. Two balls of charge q1 and q2 initially
have a velocity of the same magnitude and
direction. After a uniform electric field
has been applied during a certain time, the
direction of the velocity of the first ball
changes by 60°, and the velocity magnitude
is reduced by half. The direction of the
velocity of the second ball changes thereby
by 90°.
In what proportion will the velocity of
the second ball change? Determine the magnitude of the charge-to-mass ratio for the
second ball if it is equal to kl for the first
ball. The electrostatic interaction between
the balls should be neglected.
3.17. Small identical balls with equal
charges are fixed at the vertices of a right
1977-gon with side a. At a certain instant,
one of the balls is released, and a sufficiently long time interval later, the ball adjacent
to the first released ball is freed. The kinetic energies `of the released balls are found to
differ by K at a sufficiently long distance from
the polygon.
Determine the charge q of each ball.
3.18. Why do electrons and not ions cause
3. Electricity and Magnetism 79
collision ionization of atoms although both
charges acquire the same kinetic energy
.mv”/2 = e Arp (e is the charge of the parti`cles, and -Aq> is the potential difference)`in
an accelerating field? Assume that an atom
to_ be ionized and a particle impinging on
"it have approximately the same velocity after the collision.
3,19. Two small identical balls lying on a
horizontal plane are connected by a weight`less spring. One ball is iixed at point O
and the other is free. The balls are charged
identically, as a result of which the spring
length increases twofold.
T Determine the change in the frequency
.of harmonic vibrations of the system.
#3.20. Two small balls having the same mass
and charge and located on the same vertical at heights hl and hz are thrown in the
same direction along the horizontal at the
qsame velocity v. The first ball touches the
Xiground at a distance Z from the initial vertical.
At what height H 2 will the second ball be
§at this instant? The air drag and the effect
Jof the charges induced on the ground should
be neglected.
3.21. A hank of uninsulated wire consisting
jof sevenand a half turns is stretched between
Qtwo nails hammered into a board to which
ihe ends of the wire are fixed. The resistance
of the circuit between the nails is deterQnined with the help of electrical measuring
fnstruments.
Determine the proportion in which the
80 Aptitude Test Problems in Physics
resistance will change if the wire is unwound so that the ends remain to be fixed
to the nails.
3.22. Five identical resistors (coils for hot
plates) are connected as shown in the diagram in Fig. 80.
Fig. 80
What will be the change in the voltage
across the right upper spiral upon closing
the key K?
3.23. What will be the change in the resistance of a circuit consisting of iive identiFig. 81
cal conductors if two similar conductors are
added as shown by the dashed line in
Fig. 8i?
3;24. A wire frame in the form of a tetrahedron ADCB is connected to a d.c. source
3. Electricity and Magnetism 81
(Fig. 82). The resistances of all the edges of
the tetrahedron are equal.
Indicate the edge of the frame that
should be eliminated to obtain the maximum
p
A.
1 "" 0
6
1
Fig. 82
change in the current AI max in the circuit,
neglecting the resistance of the leads.
3.25. The resistance of each resistor in the
0
AI
· _ _ _ .13
8
. I //2
0
Rig. 83
dlrcuit diagram shown in. Fig. 83 is the same
Psnd equal to R. The voltage across the ter1-plllinals is U.
82 Aptitude Test Problems in Physics
Determine the current I in the leads if
their resistance can be neglected.
3.26*. Determine the resistance RAB between pointsA and B of the frame formed by
nine identical wires of resistance R each
(Fig. 84).
RA0
” ¤ '”
A AP 0
Fig. 84
3.27. Determine the resistance RAB between points A and B of the frame made of
thin homogeneous wire (Fig. 85), assuming
Al _a
E
1
|-—-~·—~————L—~———-+
Fig. 85
that the number of successively embedded
equilateral triangles (with sides decreasing
by half) tends to infinity. Side AB is equal
to a, and the resistance of unit length of the
wire is p.
3. Electricity and Magnetism 83
3.28. The circuit diagram shown in Fig. 86
consists of a very large (infinite) number
of elements. The resistances of the resistors
in each subsequent element differ by a facA P, 0 AP, |-—E
R2 AP2 :
}___6p{
Fig. 86
tor of k from the resistances of the resistors
in the previous elements.
Determine the resistance RAB between
points A and B if the resistances of the first
element are R1 and R,.
3.29. The voltage across a load is controlled
by using the circuit diagram shown in Fig. 87.
The resistance of the load and of the
0 P II
Fig. 87
potentiometer is R. The load is connected
to the middle of the potentiometer. The input voltage is constant and equal to U.
Determine the change in the voltage
across the load if its resistance is doubled.
5•
84 Aptitude Test Problems in Physics
3.30. Given two different ammeters in
which the deflections of the pointers are
proportional to current, and the scales are
uniform. The first ammeter is connected to
a resistor of resistance R1 and the second to
a resistor of unknown resistance Rx. At
first the ammeters are connected in series
between points A and B (as shown in
P7 PI A
Z--c:>—@——¤;ii~-( )—;
Fig. 88
Fig. 88). In this case, the readings of the
ammeters are nl and nz. Then the ammeters
are connected in parallel hetween A and I?
"°’ o
AB
. pg ·
-Q
Fig. 89
(as shown in Fig. 89) and indicate zz; and
Determine the unknown resistance Rx
of the second resistor.
3.31. Two wires of the same length but of
different square cross sections are made
from the same material. The sides of the
cross sections of the first and second wires
are dl =—- 1 mm and dg = 4 mm. The current required to fuse the first wire is I1 =
10 A. `
3. Electricity and ltlagnetism 85
Determine the current I 2 required to
fuse the seconel wire, assuming that the
amount of heat! dissipated to the ambient
per second obeys the law Q =kS (T — T am),
where S is the cross-sectional area
of the wire, T is its temperature, Tam is
the temperature of the ambient away from
the wire, and lc is the proportionality factor
which is the same for the two wires.
.3.32. The key K in circuit diagram shown
in Fig. 90 can be either in position 1 or 2.
I (K
x R·¤
._ g-! _
87
0
Fig. 90
The circuit includes two d.c. sources, two
resistors, and an ammeter. The emf of one
source is gl and of the other is unknown.
The internal resistance et the sources
should be taken as zero. The resistance of the
resistors is also unknown. One of the resistors has a varying resistance chosen in
such a way that the current through the amriteter is the same for both positions of the
liey. The current is measured and is found
to be equal to I.
Determine the resistance denoted by R,
in the diagram.
86 Aptitude Test Problems in Physics
3.33. Can a current passing through a resistor be increased by short—circuiting one
of the current sources, say, the one of emf
l " ...l
—LZJ———Fig. 91
E2 as shown in Fig. 9i? The parameters of
the elements of the circuit should be assumed to be specified.
3.34*. A concealed circuit (black box) consisting of resistors has four terminals (Fig. 92).
If a voltage is applied between clamps
1J
2I4
92
1 and 2 when clamps 3 and 4 are disconnected, the power liberated is N1 = 40 W,
and when the clamps 3 and 4 are connected,
the power liberated is N, == 80 W. If the
same source is connected to the clamps
3 and 4, the power liberated in the circuit
when the clamps 1 and 2 are disconnected
is N3 = 20 W.
l- Determine the power N4 consumed in
the circuit when the clamps 1 aud 2 are
3. Electricity and Magnetism 87
connected and the same voltage is applied
between the clamps 3 and 4.
3.35. Determine the current through the
battery in the circuit shown in Fig. 93
6K
|—-——/
01
. Pr A R2 02 0
R
Fig. 93
(1) immediately after the key K is closed and
(2) in a long time interval, assuming that
the parameters of the circuit are known.
3.36. The key K (Fig. 94) is connected in
turn to each of the contacts over short iden87 Rf
6 K ;L
IZ
Fig. 94
tical time intervals so that the change in
the charge on the capacitor over each conn_ection is small.
What will be the final charge qt on the
capacitor?
88 Aptitude Test Problems in Physics
3.37. A circuit consists of a current source
of emf é?° and internal resistance r, capacitors of capacitance C1 and C2, and resistors
of resistance R1 and R2 (Fig. 95).
6} pl 62
@2 `
8, |;·m___+_ ·
Fig. 95
Determine the voltages U1 and U2
across each capacitor.
3.38*. A perfect voltmeter and a perfect
ainmeter are connected in turn between
A
.. FK
.5
6
Fig. 96
points E and F of a circuit whose diagram is
shown in Fig. 96. The readings of the instruments are UO and I0.
Determine the current Imthrough the reSiSt01‘ of 1‘GSiSi3&DOB R COHHBOEGCI b9tW€9H
points E and F.
3. Electricity and Magnetism 89
3.39. A plate A of a parallel-plate capacitor
is Hxed, while a plate B is attached to the
wall by a spring and can move, remaining
parallel to the plate A (Fig. 97). After the
key K is closed, the plate B starts moving
and comes to rest in a new equilibrium position. The initial equilibrium separation d
between the plates decreases in this case
by 10%.
What will be the decrease in the equilibrium separation between tl1e plates if the
key K is closed for such a short time that
the plate B cannot be shifted noticeably?
:1
A6;34
E4
K/0/I 1 2
Fig. ev Fig. ea
3.40. The circuit shown in Fig. 98 is made
of a homogeneous wire of constant cross
section.
Find the ratio Q12/Q3. of the amounts of
heat liberated per unit time in conductors
1-2 and 3-4.
3.41. The voltage between the anode and
the cathode of a vacuum-tube diode is U ,
and the anode current is I.
—90 Aptitude Test Problems in Physics
Determine the mean pressure pm of electrons on the anode surface of area S.
3.42. A varying voltage is applied to the
clamps ABf(Fig. 99) such that the voltage
AH1Té
60-——-——--—— —·—--——<>D
Fig. 99
across the capacitor plates varies as shown
in Fig. 100.
Plot the time dependence of voltage across
the clamps CD.
UI
II
I _I
II
II—
Fig. 100
3.43. Two batteries of emf {E1 and E2, a
capacitor of capacitance C, and a resistor
xr xr
Br lg 0
L4; I
Fig. 101
of resistance R are connected in a circuit as
shown in Fig. 101.
3, Electricity and Magnetism 91
Determine the amount of heat Q liberatgd in the resistor after switching the key
K.
3.44. An electric circuit consists of a current source of emf 8 and internal resistance
r , and two resistors connected in paral.|.8·'°
e #2
Fig. 102
·lel to the source (Fig. 102). The resistance
R1 of one resistor remains unchanged, while
the resistance R2 of the other resistor can
be chosen so that the power liberated in .this
resistor is maximum.
Determine the value of R2 corresponding
to the maximum power.
3.45. A capacitor of capacitance C1 is discharged through a resistor of resistance R.
6}
I ik ig j|.]'6`2
Fig. 103
When the discharge current attains the
value I 0, the key K is opened (Fig. 103).
92 Aptitude Test Problems in Physics
Determine the amount of }1eat Q liberated in the resistor starting from this moment.
3.46. A battery of emf 2E, two capacitors of
capacitance C1 and C2, and a resistor of rexv ,<>
i`i””"°`i`@i 1
- 8 :0
Fig. 104
sistance R are connected as shown in
Fig. 104.
Find the amount oi heat Q liberated in
the resistor after the key K is switched.
3.47. In the circuit diagram shown in
Fig. 105, the capacitor of capacitance C
R7
j—[;l’°1
¥ xr
l0
Fig. 105
is` uncliarged when the key K is open. The
key is closed over some time during which
tht-3 C3p&Oll»0I‘ IJBCOIIIBS charged to 3 V0lt&g`0
UI
3. Electricity and Magnetism 93
Determine the amount of heat Q2 liberated dujiéingf thlis tirr;e ip tlgle resistor of éesisté
fil ance 2 i t e em o t esource is , an
its internal resistance can be neglected.
@3.48. A jumper of mass m can slide without
friction along parallel horizontal rails separated by a distance d. The rails are connected to a resistor of resistance R and
B
/°
if .2...
rpg. 106
§Iaced in a vertically uniform magnetic field
induction B. The jumper is pushed at a
velocity vo (Fig. 106).
iii Determine the distances covered by the
jumper before it comes to rest. How does
?i§’the direction of induction B affect the anf§j{ swer?
3.49. What will be the time dependence of
the reading of a galvanometer connected to
the circuit of a horizontal circular loop
Erhhe? a charged ball falls along the axis of
il-. 4 O OOP _
3..50. A small charged ball suspended on an
inextensible thread of length l moves in a
uniform time-independent upward magnetfield of induction B. The mass of the ball
**‘a gz, the cgiarge is q, andthe period of revoiu ion IS. . s
94 Aptitude Test Problems in Physics
Determine the radius r of the circle in
which the ball moves if the thread is always
stretched.
3.51.'_A metal ball of radius r moves at a
constant velocity v in a uniform magnetic
field of induction B.
Indicate the points on the ball the potential difference between which has the maximum value Acpmax. Find this value, assuming that the direction of velocity forms
an angle ot with the direction of the magnetic induction.
3.52. A direct current flowing through the
winding of a long cylindrical solenoid of
radius R produces in it a uniform magnetic
field of induction B. An electron flies into
the solenoid along the radius between its
turns ‘(at right angles to the solenoid axis).:::!·
..2., ·'·.§i.·“·.i'£§
‘!.!!;,
°=!!=•
·•n¤v·
B
Fig. 107
at a velocity v (Fig. 107). After a certain
time, the electron deflected by the magnetic
field leaves the solenoid. _
Determine the time t during which the
electron moves in the solenoid.
3.53. A metal jump-er of mass ri; can slide
without friction along two parallel metal
3. Electricity and Magnetism 95
guides directed at an angle oz to the horizontal and separated by a distance b. The guides
are connected at the bottom through an
uncharged capacitor of capacitance C, and
the entire system is placed in an upward
at all .
elf /
0 /G/.¢
Fig. 108
magnetic field of induction B. At the initial moment, the jumper is held at a distance
l from the foot of the "hump" (Fig. 108).
Determine the time t in which the released jumper reaclies the foot of the___hump.
What will be its velocity vt at the foot?
The resistance oi .the guides andffthe jumper
should be- neglected.
3.54*. A quadratic undeformable superconducting loop of mass m and side a lies in a
horizontal plane in a nonuniform magnetic
field whose induction varies in space according to the law Bx = ——-ocx, By = O,
B, = ocz -|— B0 (Fig. 109). The inductance
of the loop is L. At the initial moment, the
centre of the loop coincides with the ori~
gin O, and its sides are parallel to the :2:and y-axes. The current in the loop is zero,
and it is released.
96 Aptitude Test Problems in Physics
How will it move and where will it be
in time t after] the beginning of motion?
z
(27
5/
Fig. 109
3.55. A long cylindrical coil of inductance
L, is wound on a bobbin of diameter D1.
The magnetic induction in the coil connected to a current source is B2. After rewinding the coil on a bobbin of diameter D2,
its_ inductance becomes L2.
Determine the magnetic induction _B2
of the field in the new coil connected to the
same current source, assuming that the
length of the wire is much larger than that
of the coil.
3.56*. Two long cylindrical coils with uniform windings of the same length and nearly
the same radius have inductances L2 and
L2_. The coils are coaxially inserted into
each other and connected to a current. sou_ce
as shown in Fig. 110. The directions of
the current in the circuit and in the turns
are shown by arrows.
Determine the inductance L of such a
composite coil.
3. Electricity and Magnetism 97
3.57. A winch is driven by an electric motor with a separate excitation and fed from
a battery of emf 8 = 300 V. The rope and
the hook of the winch rise at a velocity
v1 =- 4 m/s without a load and at a velocity
v, = 1 m/s with a load of mass m == 10 kg.
[ I Mm E Cathode
~/27V To
Fig. 110 Fig. 111
Determine the velocity v' of the load and
its mass m' for which the winch has the
maximum power, neglecting the mass of the
rope and the hook.
3.58. A perfect diode is connected to an a.c.
circuit (Fig. 111).
Determine the limits within which the
voltage between the anode and the cathode
varies.
3.59. A capacitor of unknown capacitance,
a coil of inductance L, and a resistor of re42
5
Fig. 112
sistance R are connected to a source of a.c.
voltage E == go cos wt (Fig. 112). The current in the circuit is I == (ESO/R) cos cot.
7 -0771
98 Aptitude Test Problems in Physics
Determine the amplitude U 0 of the voltage across the capacitor plates.
3.60. Under the action of a constant voltage U, a capacitor of capacitance C = 10"11 F
c
1"_""l 0
UP5L0
1.
Fig. 113 Fig. 114
included in the circuit shown in Fig. 113
is charged to q1 = 10*** C. The inductance
of the coil; is L = 10"° H, and the resistance of the resistor is R = 100 Q.
Determine the amplitude qu of steadystate oscillations of the charge on the
capacitor at resonance if the amplitude of
the external sinusoidal voltage. is U 0 = U.
3.61. A bank of two series-connected capacitors of capacitance C each is charged to a
voltage U and is connected to a coil of inductance L so that an oscillatory circuit
(Fig. 114) is formed at the initial moment.
After a time 1:, a breakdown occurs in one of
the capacitors, and the resistance between
its plates becomes zero.
Determine the amplitude qc of charge oscillations on the undamaged capacitor.
3.62. How can the damage due to overheating the coil of a superconducting solenoid
be avoided?
4. Optics
4.1. A point light source S is on the axis of
a hollow cone with a mirror inner surface
(Fig. 115). A converging lens produces on a
Screw:
Fig. 115
SG1‘00I1 13110 image of the S0l1I‘C0 f01‘I[10d by 13110
PHYS l1I1d0I‘g0lIlg El SlI1gl0 I‘0fl0C»tlOl’l 313 th0
inner surface (direct rays from the source
do not fall on the lens).
Cs
<\
. gw
Fig. ue Fig. m
What will happen to the image if the lens
is covered by diaphragms like those shown
in Figs. 116 and 117?
7t
100 Aptitude Test Problems in Physics
4.2. Remote objects are viewed through a
converging lens with a focal length F ==
9 cm placed at a distance a = 36 cm in
front of the eye.
Estimate the minimum size of the screen
that should be placed behind the lens so
that the entire field of view is covered.
Where should the screen be placed? Assume
that the radius r of the pupil is approximately 1.5 mm.
4.3*. A cylindrical transparent vessel of
height h (h <RvcS), where Rm, is the radius of the vessel, is filled with an ideal gas
of molar mass p. at a temperature T and a
pressure pi,. The dependence of the refractive index n of the gas on its density p
obeys the law n = 1 -1- ocp. The vessel is
rotated at an angular velocity to about its
axis. A narrow parallel beam of light of radius rbcam is incident along the axis of the
vessel.
Determine the radius R of the spot on the
screen placed at right angles to the vessel
axis at a distance L behind the vessel, assuming that the change in the gas pressure
at each point of the vessel due‘to rotation
is small as compared with po. The effect of
the end faces of the vessel on the path of
the rays should be neglected.
4.4. A telescope with an angular magnification k = 20 consists of two converging
thin lenses, viz. an objective with a focal
length F = 0.5 m and an eyepiece which
can be adjusted to the eye within the limits
between D__ = —·7 D and D+ = +10 D
4. Optics 101
(during the adjustment, the eyepiece is
displaced relative to the objective).
What is the smallest possible distance a
from the objective starting from which
remote objects can be viewed by an unstrained normal eye through this telescope?
4.5. Can a diverging lens be used to increase the illuminance of some regions on
the surface of a screen?
4.6. A point light source is exactly above
a pencil erected vertically over the water
surface. The umbra of the pencil can be
seen at the bottom of the vessel with water.
If the pencil is immersed in water, the size
of the dark spot at the bottom increases.
When the pencil is drawn out of water, a
bright spot appears instead of the dark one.
Explain the described phenomena.
4.7. If an illuminated surface is viewed
through the wide hole in the body of a ballpoint pen, several concentric dark and
bright rings are seen around the narrow hole
in the body.
Explain why these rings are observed.
4.8. A point light source S is at a distance
I == 1 m from a screen. A hole of diameter
sd = 1 cm, which lets the light through, is
a-made in the screen in front of the light
lsource. A transparent cylinder is arranged
¤—between the source and the screen (Fig. 118).
The refractive index of the cylinder materinal is n = 1.5, the cylinder length is l = 1 m,
and the diameter is the same as that of the
102 Aptitude Test Problems in Physics
What will be the change in the luminous
flux through the hole? Neglect the absorption of light in the substance.
4.9. The objective and the eyepiece of a
telescope are double—convex symmetrical
lenses made of glass with a refractive index
Ycreen
S[_`Q’;;`] rl
iz
Fig. 118
ng, = 1.5. The telescope is adjusted to
infinity when the separation between the
objective and the eyepiece is L0 == 16 cm.
Determine the distance L separating the
objective and the eyepiece of the telescope
adjusted to infinity with water poured in
the space between the objective and the eyepiece (nw = 1.3).
4.10. A spider and a fly are on the surface
of a glass sphere. Where must the fly be
for'the spider to be able to see it? Assume
that the radius of the sphere is much larger than the sizes of the spider and the fly.
The refractive index of glass is ng! =-· 1.43.
4.11. A point light source S is outside a cylinder on its axis near the end face (base).
4. Optics .103
Determine the minimum refractive index
n of the cylinder material for which none of
the rays entering the base will emerge from
the lateral surface.
4.12. Two converging lenses are mounted
at the ends of a tube with a blackened inner
lateral surface. The diameters of the lenses
are equal to the diameter of the tube. The
focal length of one lens is twice that of the
other lens. The lenses are at such a distance
from each other that parallel light rays
incident along the axis of the tube on one
lens emerge from the other lens in a parallel
beam. When a wide light beam is incident
on the lens with the larger focal length, a
bright spot of illuminance E1 is formed on
the screen. When the tube is turned through
180°, the bright spot formed on the screen
has an illuminance E2.
Determine the ratio of illuminances on
the screen.
4.13. An amateur photographer (who is an
expert in geometrical optics) photographs
the facade of a building from a distance of
100 m with a certain exposure. Then he
decides to make a photograph from a distance of 50 m (to obtain a picture on a larger scale). Knowing that the area of the
image will increase by a factor of four, he
decides to increase accordingly the exposure
in the same proportion. After developing
the film, he finds out that the first picture
is good, while the exposure for the second
photograph has been "'chosen incorrectly.
Determine the factor by which the expo-
104 Aptitude Test Problems in Physics
sure had to be changed for obtaining a good
picture and explain why.
4.f-4*. Aifisherman lives on the shore of a
bay forming a wedge with angle on in a house
located at point A (Fig. 1.19). The distance
D
\ d B•
A. h_ G
Fig. 119
from point A to the nearest point C of the
bay is h, and the distance from point D
to point A is l. A friend of the fisherman
lives across the bay in a house located at
point B. Point B is symmetric (relative to the
bay) to point A. The fisherman has a boat.
Determine the minimum time t required
for the fisherman to get to his friend from
the house provided that he can move along
the shore at a velocity v and row the boat
at half that velocity (n = 2).
4.f5. The image of a point source S' lying
at a distance b from a transparent sphere is
formed by a small diaphragm only by rays
close to the optical axis (Fig. 120).
Where will the image be after the sphere
is cut into two parts perpendicular to the
horizontal axis, and the plane surface of
the left half is silvered?
4. Optics 105
4.16. A glass porthole is made at the bottom
of a ship for observing sealife. The hole
diameter D = 40 cm is much larger than
the thickness of the glass.
s __ 6"
-.°-.-—·•
®J
Fig. 120
Determine the area S of the iield of vision at the sea bottom for the porthole if
the refractive index of water is Nw = 1.4,
and the sea depth is h = 5 m.
4.17*. Let us suppose that a person seating
opposite to you at the table wears glasses.
Can you determine whether he is shortsighted or long-sighted? Naturally, being
pa polite person, you would not ask him to
let you try his glasses and in general would
make no mention of them.
4.18. A person walks at a velocity v in a
straight line forming an angle ot with the
plane of a mirror.
Determine the velocity vm at which he
approaches his image, assuming that the
object and its image are symmetric relative to the plane of the mirror.
4.19. Two rays are incident on a spherical
mirror of radius R =—-· 5 cm parallel to its
optical axis at distances hl = 0.5 cm and
hz { 3 cm.
106 Aptitude Test Problems in Physics
Determine the distance Ax between the
points at which these rays intersect the optical axis after being reflected at the mirror.
4.20. The inner surface of a cone coated by a
reflecting layer forms a conical mirror. _A
thin incandescent filament is stretched in
the cone along its axis.
Determine the minimum angle on of the
cone for which the rays emitted by the filament will be reflected from the conical surface not more than once.
Solut1ons
1. Mechanics
1.1. Let us first assume that there is no friction.
Then according to the energy conservation law, the
velocity v of the body sliding down the inclined
plane from the height h at the foot is equal to the
velocity which must be imparted to the bod for its
ascent to the same height h. Since for a bocfy moving up and down an inclined plane the magnitude
of acceleration is the same, the time of ascent will
be equal to the time of descent.
I , however, friction is taken into consideration,
the velocity v1 of the body at the end of the descent
is smaller than the velocity v (due to the work done
against friction), while the velocity v2 that has to
be imparted to the bod for raising it along the
inclined plane is larger than v for the same reason.
Since the descent and ascent occur with constant
(although different) accelerations, and the traversed
path is the same, the time tl of descent and the
time tz of ascent can be found from the formulas
viii _ Vat:
where s is the distance covered along the inclined
plane. Since the inequality v1 < vg is satisfied, it
ollows that t > t2. Thus, in the presence of slidfriction, the time of descent from the height h
is sonier than the time of ascent to the same height.
W ile solving the problem, we negtlected an
air drag. Nevertheless, it can easily be s own that
if an air drag is present in addition to the force of
gravity and the normal reaction of the inclined
plane, the time of descent is always longer than the
108 Aptitude Test Problems in Physics
time of ascent irres ective of the type of this
force. Indeed, if in the process of ascent the body
attains an intermediate height h', its velocity v'
at this point, required to reach the height h in
the presence of drag, must be higher than the velocity in the absence of drag since a fraction of the
kinetic energy will be trans ormed into heat during
the subsequent ascent. The body sliding down
from the height h and reaching the height h' will
have (due to the work done by the drag force) a
velocity v" which is lower than the velocity of the
body movin down without a drag. Thus, while
passing bg Sie same point on the inclined plane,
the ascen ing body has a higher velocity than the
descending body. For this reason, the ascending
body will cover a small distance in the vicinity of
point h' in a shorter time than the descending body.
Dividing the entire path into small regions, we
see that each region will be traversed by the ascending body in a shorter time than by the descending body. Consequently, the total time of ascent
wi 1 be shorter than the time of descent.
1.2. Since the locomotive moves with a constant
deceleration after the application of brakes, it will
come to rest in t =. v/ a = 50 s, during which it will
cover a distance s == v“/(2a) = 375 m. Thus, in
1 min after the application of brakes, the locomotive will be at a distance l = L -— s == 25 m
from the traffic light.
1.3. At the moment the pilot switches off the engine, the helicopter is at an altitude h = at?/2.
Since the sound can no longer be heard on the
ground after a time t2, we obtain the equation
2
t2 = *1+% •
where on the right-hand side we have the time of
ascent-of the helicopter to the altitude h and the
time taken for the sound to reach the ground from
the altitude h. Solving the obtained quadratic
equation, we End that
c. 2 c c
*1:}/(Ti)+22"=‘“;·
Solutions 109
We discard the second root of the equation since
it has no physical meaning.
The velocity v of the helicopter at the instant
when the engine is switched off can be found from
the relation
”=‘"*=“ )2+2%' '*"%`;|
= `|/c2—|—2act2-—c=80 mls.
1.4. During a time tl, the point mass moving with
an acceleration a will cover a distance s = at}/2
and will have a velocity v = atl. Let us choose the
a:-axis as shown in Fig. 121. Here point 0 marks
S=at,2/Z
O A ,1;
Fig. 121
the beginning of motion, and A is the point at
which the body is at the moment tl. Taking into
account the sign reversal of the acceleration and
applying the formula for the path length in uniform y varying motion, we determine the time t2
ia which the ody will return from point A to
point O:
atf at§
°‘=T+“‘1‘2"‘2* ·
whence t2 = tl (1 -|—
The time elapsed from the beginning of motion
to the moment of return to the initial position can
be determined from the formula
¥= t1+tz= t1(2`|"`
1.5., We shall consider the relative motion of the
bodies from the viewpoint of the first body. Then
at the initial moment, the first body is at rest (it
can be at rest at the subsequent instants as well),
110 Aptitude Test Problems in Physics
while the second body moves towards it at a velocity vl + v2. Its acceleration is constant, equal
in magnitude to al -|·· az, and is directed against
the initial velocity. ‘The condition that the bodies
meet indicates that the distance over which the
velocity of the second body vanishes must belonger than the separation between the bodies at the
beginning of motion; hence we obtain
I I =___ (vi-!-v¤)“
ma 2 (*11+ ¤¤) '
1.6. Since the balls move along the vertical, we
direct the coordinate axis vertically upwards. We
plot the time dependence of the projections of the
velocities of the balls on this axis. Figures 122
D7 L°;
up + ` lll M
k\E
`"2"r
"vh · ·»l·i·· — wl-}- ·· ·+·
Fig. 122
UZ tz
fg - .._......
2
va _ if
Fig. 123
and 123 show the dependences v1 (t) and v, (t)
respectively (the moments of the beginning of motion are not matched so far). These graphs present
infinite sets of straight line segments with e ual
slopes (since the acceleration is the same). These
Solutions 111
segments are equidistant on the time axis, their
separations being tl = 2]/2hl/g for the first hall
and tz = 2]/2h,/g for the second ball. Since
hl = 4h, by hypothesis, tl = 2t , i.e. the frequency ofamotion of the second ball is twice as
high as that of the first ball. It follows from the
ratio of the initial heights that the maximum velocities attained by the balls will also differ by a
factor of two (see Figs. 122 and 123):
v1IIl8X"= 2v2m8X= l/2h18’= VoThere are two possibilities for the coincidence
of the velocities of the balls in magnitude and direction. The velocities of the balls may coincide
for the first time 1: =·- ntl s after the beginning of
motion (where n = 0, 1, 2, . . .) during the time
interval tl/4, then they coincide 3tl//1 s after the
beginning of motion during the time interval t /2.
Su sequently, the velocities `will coincide with a
period tl during the time interval tl/2. The other
possibility consists in that the second ball starts
moving 1: =.= tl/2 + nt, s (where n = 0, 1, 2, . . .)
after the first ball. After tl/4 s, the velocities of
the balls coincide for the first time and remain identical during the time interval tl/2. Subsequently,
the situation is repeated with a period tl.
For other starting instants for the second ball,
the velocity graphs will have no common points
ugion superposition because of the multiplicity of
t e periods of motion of the balls, and the problem
will have no solution.
1.7*. Let us consider the motion of a ball falling
T freely from a height H near the symmetry axis starting from the moment it strikes the surface. At the
‘moment of impact, the ball has the initial velocity
vo =· ]/2gH (since the impact is perfectly elastic),
grand the direction of the velocity vo forms an angle
with the vertical (Fig. 124).
Let the displacement of the ball along the horiggontal in time t after the impact be s, Then
[ig, sin 2a-t = s. Hence we _obtain t = s/(]/ 2gH ><
km 2oz), where vo sin 2ot is the horizontal com-
112 Aptitude Test Problems in Physics
ponent of the initial velocity of the ball (the ball
does not strike the surface any more during the
time t). The height at which the ball will e in
time t is
g¢’
Ah=h0+U0 COS 2G•Z·---2- ,
where vo cos 2a is the vertical component of the
initial velocity of the ball.
, Since the ball starts falling from the height H
near the symmetry axis (the angle on is small), we
can assume that ho z 0, sin 2oz z 2cz, cos 2oz z 1,
I
I
I
I
I
I
I
2&"° {
. ·¤° · —
.`/
Fig. 124
and s as Roz. Taking, into account these and other
relations obtained a ove, we End the condition for
the ball to get at the lowest point on the spherical
surface:
... ....2..... = _j._
1/ I2?] sin 2a. 2 ’
,.. ¤¢’ -£_... R’ Ah ~v0t—· 2 —· 2 %·I?———0,
Hence H = R/8.
1.8. Since the wall is smooth, the impact against
the wall does not alter the vertical component of
the ball velocity. Therefore, the total time tl of
Solutions 113
motion of the ball is the total time of the ascent
and descent of the body thrown upwards at a velocity va sin on in the gravitational field. Consequentl , t = 2v,, sin cc/g. The motion of the ball along
the horizontal is the sum of two motions. Before
the collision with the wall, it moves at a velocity
vo cos ot. After the collision, it traverses the same
distance backwards, but at a different velocity.
In order to calculate the velocity of the backward
motion of the ball, it should be noted that the velocity at which the ball approaches the wall (along
the horizontal) is vo cos cz + v. Since the impact
is perfectly elastic, the ball moves away from the
wa l after the collision at a velocity vo cos a + v.
Therefore, the ball has the following horizontal
velocity relative to the ground:
(vocosa-|— v) -|— v= v0cosa—|— 2v.
If the time of motion before the impact is t, by
equating the distances covered by the ball before
and after the collision, we obtain the following
equation:
vo cos 0:.-1 = (tl —— t) (vo cos on —|— 2v).
Since the total time of motion of the ball is tl =
2v0 sin cx./g, we find that
t__ vo sin cc (vo cos cz-}- 2v)
— g (vo cos cz + v) °
i.9*. Figure 125 shows the top view of the trajectorg of the ball. Since the collisions of the ball with
t e wall and the bottom of the well are elastic, the
magnitude of the horizontal component of the
ball velocity remains unchanged and equal to v.
The horizontal distances between points of two
successive collisions- are AAI = AIA, = A,A, =
i. . . =—= 2r cos ot. The time between two successive
collisions of the ball with the wall of the well is
i til = 2r cos cx/v.
,, t` The vertical component of the ball velocity
htdocs not change upon a collision with the wall and
wverses its sign upon a collision with the bottom.
.-0771
114 Aptitude Test Problems ih Physics
The magnitude of the vertical velocity component
for the first impact against the bottom is L/2gH,
and the tnme of motion from the top to the ottom
of the well is tz = ]/2H/g.
"2
Q
ZG \\ X
P \\\ A w
¢z,’
is
A1
Fig. 125
Figure 126 shows the vertical plane development of the polyhedron AIA A3. . . . On this
development, the segments of the trajectory of the
ball Inside the well are parabolas (complete paraA ___I ___ I_ _I ___ I___ '*j¤__~g_I_
I- -1 As T I I
I·II(
IIII
°= A4 I I
III
III
III
I2r·c0s:d
I·*·——>I
Fig. 126
bolas are the segments of the trajectory between
successive impacts against the bottom). The ball
can _“get out" of the well if the moment of the
maximum ascent along the parabola coincides with
the moment of an impact against the wall (i.e. at
the moment of maximum ascent, the ball is at
point An of the well edge). The time tl is connected
to the time tz through the following relation:
7;.9*1 “= 2kt2I
Solutions 115
where n and k are integral and mutually prime
numbers. Substituting the values of tl and t,, we
obtain the relation between v, H, r, and oz. for
which the ball can "get out" of the well:
nr cosot = k 1/2-___I_1l
v8°
1.10. From all possible trajectories of the shell,
we choose the one that touches the shelter. Let us
analyze the motion of the shell in the coordinate
system with the axes directed as shown in Fig. 127.
u
C,
5/ ·:·_ Kg} m
\
AMt
6 ,4 `”"""**·····
lm. .;...{._-_
Fig. 127
The "horizontal" component (along the axis Ax) of
the initial velocity of the shell in this system is
.v,,_, = vo cos (T -— oc), and the "vertical" component (along t e axis Ay) is vw = v,,,sin (qa —- u),
where qa is the angle formed by the direction ;_of the
initial velocity of the shell.and the horizon tal.
Point C at which the trajectory of the shell
if/touches the shelter determines the maximum altiytude h' of the shell above the horizontal. Fig'illl'9 127 shows that h' =l sin oc. The projection of
lthe total velocity v oi the shell on the axis A y is
jgero at this point, and
ln; 2g' ’
116 Aptitude Test Problems in Physics
where g' = gicos oz. is tbe_],"free-fall" acceleration
in the coordinate system zA y. Thus,
vf, sinz (cp — cc) = 2gl cos on sin oc.
Hence it follows, in particular, that if
vg < 2gl cos on sin oc = gl sin 20:.
by hypothesis, none of the shell trajectories will
touch the shelter, and the maximum range Lmax
will correspond to the shell fired at an angle cp =
:1/4 to the horizontal. Here Lmax = vg/g.
I.
v§ _> gl sin 2a
by hypothesis, the angle at which the shell should
be fired to touch the shelter will, he
,:,,;.0,+ ,,,, in
vu
If, moreover, the inequality
v§ .
U-——f $s1n2cx
is satisfied by hypothesis, which means that the
condition cpt > rc/4 holds (prove thisl), the angle
cp at which miie shell having the maximum range
should be tired will he equal to at/4, and Lmax =
vg/g. If, however, the inverse inequality
v§ . 2
—-————v% +281 >s1n on
is known to be valid, which in turn means that
cpt < at/4, we have
l°2
cp-—=<pt==o¤—|—arcsin ,
v2 .
Llnax-:`—*;)*
zig. mz (a—|—arcsin -..-.-l/"ji“ 2°‘
o
Solutions 117
1.11. Let us su pose that hail falls along the
vertical at a velocity v. In the reference frame
fixed to the motorcar, the angle of incidence of
hailstones on the windscreen is equal to the angle
of reflection. The velocity of a hailstone before it
strikes the windscreen is v — v1 (Fig. 128). Since
3
u—u,
A1
`°"r
fi
Fig. 128
hailstones are bounced vertically ugwards (from
the viewpoint of the drive? after t e reflection,
the angle of reflection, an hence the angle of
incidence, is eqlual to 81 (81 is the slope of the
windscreen of t e motorcar). Consequently, on +
281 = n/2, and tan on = v/v1. Hence tan on =
tan (:1:/2 — 281) = cot 281, and v/v1 = cot 281.
Therefore, we obtain the following ratio of the
velocities of the two motorcars:
-2-L.- cot 2Bg -3
Us -— cot281 _— o
1.12. Let us go over to a reference frame moving
with points A and B. In this system, the velocities o points A and B are zero. Since the distances
AC and BC are constant, point C, on the one
hand, can move in a circle of radius AC with the
centre at point A, and on the other hand, in a
.circle of radius BC with the centre at point B.
Therefore, the direction of velocity of point C
must be perpendicular both to the straight line
AC and the straight line BC. Since points A, B,
and C do not lie on the same_ straight line, the
direction of velocity at point C would be perpen-
118 Aptitude Test Problems in Physics
dicular to two intersecting straight lines AC and
BC, _ which is impossible. Consequently, in the
{novmg reference tame, the velocity of point C
18 zero, while in the initial system (faxed to the
ground), the velocity of oint C is equal to the
Velocity of points A and) B.
If point C lay on the straight line AB (in the
reference frame fixed to the sheet of plywood) and
its velocity differed from zero, after a certain small
hme interval either the distance AC or the distance BC would increase, which is impossible.
Therefore, in the motion of the sheet of plywood under consideration, the velocities of all
the points are identical.
i.13. Let a car get in a small gap between two
other cars. It is parked relative to the pavement
as shown in Fig. 129. Is it easier for the car to be
7 \ ry
Ur 6
é_
\rI§
. /’-· 6~'
0 :Z-,_ 6
”I
A vo
Fig. 129 Fig. 130
driven out of the gap by forward or backward motion? Since only the front wheels can be turned,
the centre O of the circle along which the car is
driven out for any manoeuvre (forward or backward) always lies on the straight line assing
through the centres of the rear wheels of the car.
Consequently, the car being driven out is more
likely to touch the hind car •during backward motion than the front car during forward motion (the
centre of the corresponding circle is shifted back-
Solutions 119
wards relative to the middle of the car). Obviously,
a driving out is a driving in inversed in time.
Therefore, the car should e driven in a small gap
by backward motion.
1.14*. Let us consider the motion of the plane
starting from the moment it goes over to the circular path (Fig. 130). By hypothesis, at the upper
point B of the path, the velocity of the plane is
v1 = vo/2, and ence the radius r of the circle
described by the plane can be found from the relation
J2-%—=v§—2a0·2r,
which is obtained from the law of motion of the
plane for h ··-= 2r. At point C of the path where the
velocity of the plane is directed upwards, the total
acceleration will be the sum of the centripetal
acceleration dc = vg/r (vg = vg — 2a,,r, where vc
is the velocity of the plane at point C) and the
tangential acceleration at (this acceleration is
responsible for the change in the magni-tude of the
velocity).
In order to iind the tangential acceleration, let us
consider a small displacement of the plane from
point C to point C' Then vg. = vg — 2a,, (r + Ah).
Therefore, vg. — vg = --2a,, Ah, where Ah is the
change in the altituile of the plane as it goes over
to point C'. Let us divide both sides of the obtained
relation bi; the time interval At during which this
change ta es place:
¤%·—·% __ _2a, Ah
At — At °
Then, making point C' tend to C and At to zero ,
wp obtain
hlatvc = -—2a0vC.
{Hence at = —a0. The total acceleration of the
gglplane at the moment when the velocity has the
120 Aptitude Test Problems in Physics
upward direction is then
v“ "'“
r3
1.15. Let tne velocity of the drops above the erson relative to the merry-go—round be at an angl)e on
to the vertical. This angle can be determined from
the_velocity triangle shown in Fig. 131.
Since in accordance with; the velocity composion rule v0=v,B]—|-vm_g_,., where vm_g_,. is the
velocity of the merry-go-round in the region of loU="”m.g,r
Ug ..
a 6, · L A
E\R 5\`
-¤ 'A§i"°°
’“ \I
Ure! °0 6 A
Fig. 131 Fig. 132
cation of the person, v -·= v — v _ _ . The velocity of the merry—go·i;gl1nd ig vm_gIg• g-; mr. Consequently cot on = v0/(cork).
Therefore, the axis 0 the umbrella should be
tilted at the angle cz = arccot [vo/(mr;} to the vertical in the direction of motion of t e merry-goround and perpendicular to the radius of the latter.
1.16*. Let the board touch the bobbin at poin: C
at a certain instant of time. The velocity of point
C is the sum of the velocity vo of the axis O of
the bobbin and the velocity of point C (relative to
point O), which is tangent to the circle at point C
and equal in magnitude to vo (since there is no
slipping). If _the angular velocity of the board at
this instant is no, the linear velocity of the point
of the board touching the bobbin will be
to}? t£111"* (cx/2) (Fig. 132). Since the board remains
Solutions 121
in contact with the bobbin all the time, the
velocity of point C relative to the board is directed along the board, whence wR tan‘1 (on/2) =
vo sin oz. Since there is no slipping of the bohbin
over the horizontal surface, we can write
.&.....L.
R _ R-}-r °
Therefore, we obtain the following expression for
the angular velocity mz
(0.: v Sinwtan iz 2v sinz (a/2)
R-|-r 2 (R—i—r) cos (oc/2) °
1.17. The area of the spool occupied by the wound
thick tape is S1 = JI (r§—r§) = 8:rtr§. Then the
length of the wound tape is l = S1/d = 8:: (r§/d),
where d is the thickness of the thick tape.
The area of the spool occupied by the wound
thin tape is S, = rt (rf — rf), where ri. is the final
radius of the wound part in the latter case. Since
the lengths of the tapes are equal, and the tape
thickness in the latter case is half that in t e
former case, we can write
2n(r'2—-1-*) l
l==····—%1····J·- , rr2——r%=4r`i·
Consequently, the final radius ri of the wound part
in the latter case is
T;= T/5 T1The numbers of turns N, and N, of the spool
for the former and latter winding can be written as
__ 2r, __ (1/5-1)r,
N1— -3* » Nz—·*·····&7i······· »
whence t, = (Vg-- 1) tl.
122 Aptitude Test Problems in Physics
1.18. Let the initial winding radius be 4r. Then
the decrease in the winding area as a result of the
reduction of the radius by half (to 2r) will be
S = sn: (16r2 — 4r2) = i2rtr2,
which is equal to the product of the length ll of
the wound tape and its thickness d. The velocity v
of the tape is constant during the operation of the
tape-recorder; hence ll = vtl, and we can write
12:tr2 = vtld. (1)
When the winding radius of the tape on the
cassette is 1·educed by half again (from 2r to r),
the winding area is reduced by rc (4r“ — rz) =
3:tr*, i.e.
3nr* = vtzd, (2)
where tz is the time-during which the winding radius will be reduced in the latter case. Dividing
Eqs. (1) and (2) termwise, we obtain
1.19. Let us go over to the reference frame fixed to
ring O' In t is system, the velocity of ring O is
vl/cos oc and is directed upwards since the thread is
inextensible, and is pulled at a constant velocity
v1 relative to ring O' Therefore, the velocity of
ring 0 relative to the straight line AA' (which is
stationary with respect to the ground) is
__ vi _ _ 2 sin? (oz/2)
v“_ cosoc vlgvl coscx
and is directed upwards.
1.20. By the moment of time t from the beginnin} of motion, the wedge covers a distance s -·=
arg 2 and acquires a velocity vwed = at. During
this time, the load will move along the wedge over
the same distance s, and its velocity relative to the
wedge is vn] = at and directed upwards along the
Solutions 123
wedge. The velocity of the load‘ relative to the
ground is vl = vm] —l— vwcd, i.e. (Fig. 133)
vi: Zvwcd sin —g—= (Za sin gi) t,
and the angle formed by the velocity vl with the
horizontal is
B= -E;-gg- = const.
Thus, the load moves in the straight line forming
the angle B = (n ·—- ot)/2 with the orizontal. The
/ "wed
grel ` Ul
‘/
fl / ‘ L \
f7 7 ‘ %.
Fig. 133
acceleration of the load on the wedge relative to
the ground is
yay = 2a sin {zi ,
1.21. The velocity of the ant varies with time
according to a nonlinear law. Therefore, the mean
velocity on different segments of the path will not
be the same, andthe we 1-known formulas for mean
velocity cannot be used here.
, Let us divide the path of the ant from point A
to point B into small segments traversed in eglual
time intervals At. Then At = Al/vm (Al), w ere
6 (Al) is the mean velocity over a given segment
This formula su gests the idea of the so ution
fof the problem: we plot the dependence of 1/ vm (Al
124 Aptitude Test Problems in Physics
on l for the path between points A and B. The
graph is a segment of a straight line (Fig. 134).
The hatched area S under this segment is numerJ
v
L ______________ ·
U2 (
7
a ······ " f
8%
0 z, zz 1
Fig. 134
ically equal to the sought time. Let us calculate
this area:
_ 1/v1+1/v, _ _ 1 1 I2)
S — 2 (lz ¢1)——( 2,,1 —|— 2,,1 ,1 (ia lr)
=__ lg-}?
2v1l1
since 1/v, = (1/v1) Z2/ll. Thus, the ant reaches
point B in the time
4 m*—1 m2
t-2X2 m/s><10"><1 m—75 S'
1.22. Obviously, all the segments of the smoke trace
move in the horizontal direction with the velocity of wind. Let us consider the path of the locomotive in the reference frame moving with the
wind (Fig. 135). Points A' and B' of the smoke
trace correspond to the smoke ejected by the locomotive at points A and B of its circular path rela-tive to the ground. Obviously}, AA' || BB'. It can
easily be seen that the pat of the locomotive
relative to the reference frame moving with the
wind is the path of the point of a wheel of radius R
rotating at a velocity vloc and moving against the
wind_(to the left) at a velocity vwmd. This trajectory 18 known as a cycloid. Depending on the ratio
Solutions 125
f velocities v and v I d the shape of _the lower
gart of the plgth will wbte ,difierent. It 18 a loop
(when vm, > vwgnd, Fig. 136),uor a ,parabola (w hen
vm < vwmd, F1g. 137), or a beak (when vlcc =
vwmd, Fig. 138). The latter case corresponds to the
WK 0 I/K \_
i 0 1 1 O' }
\Il;
` JI \\ ' rl
A;§_;ZfZ;;jZ'j.'.'j. " _Q4,_q;“//’
Fig. 135
vw vwznu Um <vwmd U’”°Tv”i”d
1i/
;I
....•-•- """‘ ·—-•- -—·—•—- ---0--.0 0/
Fig. 136 Fig. 137 Fig· 138
”v¢ ‘···· · 0,
“ "" ?~
1 $% ——-- ‘° · —————
lei "
Fig. isa
condition of our problem. Thus, the velocity of the
ggvind is vwmd = vl C = 10 m/s, and in order to {ind
Egnts direction, we draw the tangent CC" from the
E; int of the "beak" to the path of the locomotive
zifviz. the circle) relative to the ground.
Let the merry-go-round turn through a cerBIIQIB (P YVG COIlStI‘l1Ci.· 3 pOll`lt O
126 Aptitude Test Problems in Physics
(OA = R) such that points O, S, A, and J lie on
the same straight line. Then it is clear that ON =
R -|— r at any instant of time. Besides, point O is at
rest relative to John (Sam is always opposite to
John). Therefore, from J ohn’s point of view, Nick
translates in a circle of radius R + r with the centre at point O which moves relative to the ground
in a circle of radius R with the centre at point A .
From Nick’s point of view, Sam translates in a
circle of radius R -|— r with the centre at point O
which is at rest relative to Nick. However, point O
moves relative to the ground in a circle of radius r
with the centre at point B.
1.24. Since the hoop with the centre at point O
is at rest, the velocity vA of the upper point A ol
"intersection" of the hoops must be directed along
the tangent to the circle with centre O1 at any inUA
A
,¢h\
-JV
Fig. 140]
stant of time 140). At any instant of time,
the segment A ivides the distance d = OO, between the centres_of the hoops into two equal parts,
and hence the horizontal projection of the velocity
vA is always equal to v/2. Consequently, the velocity vA forms an angle cp = rt/2 —— cz w1th the
horizontal and is- given by
vA`* 2cosq> " 2sincx °i
Solutions 127
Since sin ct = 1/ 1 ·—- cos° on =·- }/1 — (d/2R)?, the
lvelocity of the upper point of "intersection" of the
oops 1S
,, ...........Y.......
A 2 i/1 —(d/2R)* °
1.25. By hypothesis the following proportion is
preserved between the lengths I1, lz, and I, of
segments A,,A1, A,,A2, and A0A3 during the motion:
llzlzzls = 3:5:6.
Therefore, the velocities of points A1, A2, and A,
are 'to one another as
and hence (Fig. 141)
__ v _ 5v
"Ar‘ 1T· "A·—"6‘·
Let us now consider the velocity] of the middle
link (A 1B,A ,02) at the instant w en the angles
/12
- A A ' A5
,
Fig. 141
of the construction are equal to 90°._In the reference
frame moving at a velocity v A1, the velocity vb_
of point B, is directed at this instant along B2A,.
The velocity of point} A is directed along the
horizontal and is given hy
I.1v
vA|:: DA]-_ UA1 K? •
128 Aptitude Test Problems in Physics
From;the condition of 'P inextensibility of the rod
B,.4 ,, it follows that
, __ , . ar: _ v }//2
UB.—UA·SlII—&——·······B···*•
We can find the velocity of point B relative to a
stationary reference frame by using the cosine law:
, 2 j/2 , __ 17
vg': vit-}- vg.-|-( --2 ) vA1vB·--jg v2,
j/17
vB’:""G'*'* v•
1.26. If the thread is pulled as shown in_Fig. 142,
the bobbin rolls to the right, rotating clockwise
about its axis.
A
zz
u
"
6* /
Fig. 142
[.For{point B, thefsum of the projections of velocitywvo of translatory motion and the linear velocity of rotary motion (with an angular velocity on)
on the direction of the thread is equal to v:
v= v0sincc— cor.
Since the bobbin is known to move over the horizontal surface without slipping, the sum of the
projections of the corresponding velocities for
point C is equal to zero:
D0 *‘ z 0•
Solutions 129
Solving__the obtained equations, we find that the
velocity vo is
,, ...._;1L._.
° — R sin oz. —r °
It can be seen that for R sin oc == r (which corres onds to the case when ploints A, B, and C lie on
the same straight line), t e expression for vo becomes meaningless. It should also be noted that the
obtained ex ression describes the motion of the
bobbin to the right (when point B is above the
straight line AC and R sin ot > rgl as well as to
the left (when point B is below t e straight line
AC and R sin ot < r).
1.27. The velocities of the points of the ingot lying on a segment AB at a given instant uni ormly
vary from v1 at point A to v2 at point B. Conse/// vi
/
r\
lx
I\
1I
\
\
1:* A J /
Fig. 143
quently, theljvelocity offpoint O (Fig. 143) at a
given instant is zero. Hence point O is an instantaneous centre of rotation. (Since the ingot is threedimensional, point O lies on the instantaneous rotational axis w ich is perpendicular to the plane of
the figure.) Clearly, at a given instant, the velocity
sv corresponds to the points of the ingot lying on
the circle of radius OA, while the velocity v, to
goints lying on the circle of radius OB. (In a threenmensiona ingot, the points having such velocities lie on cylindrical surfaces with radii OA and
OB respectively.)
-0-0771
130 Aptitude Test Problems in Physics
1.28. In order to describe the motion of the block,
we choose a reference frame fixed to the conveyer
belt. Then the- velocity of the block at the initial
moment is vb] = vo + v, and the block moves
with a constant acceleration a = —-pg. For the instant of time t when the velocity of the block
vanishes, we obtain the equation 0 = vo —|- v -—
pgt. Hence the velocity of the conveyer belt is
v= p.gt·-—v0=3 m/S.
1.29. Let- us write the equation of motion for the
body over the inclined plane. Let the instantaneous coordinate (the displacement of the top of
the inclined plane) be x; then
ma = mg sin oc —- mg cos oc·bx,
where m is the mass ofthe body, and a is its acceleration. The form of the obtained equation of motion resembles that of the equation of vibratory
motion for a body suspended on a spring of rigidity k = mg cos oc·b in the field o the "force of
gravity" mg sin oc. /This analogy with the vibratory
motion helps solve the problem.
Let us determine the position ::0 of the body
for which the sum of the forces acting on it is
zero. It will be the "equilibrium position" for the
vibratory motion of the body. Obviously,
mg sin cz ·—- mg cos ot·b:z:,, = 0. Hence we obtain
xo = (1/b) tan oc. At this moment, the body will
have the velocity vo which can he obtained from
the law of conservation of the mechanical energy of
the body:
mug -—-mg sinot a: M3
"E""' ° °""`2'
= mg sin ot·x,,-- xg,
·a
v§=2g:c0 sin ct—gb:::§ cosoc=·-%In its further motion, the body will be displaced "a ain by aio (the "amplitude" value of vibrations, which can easily be obtained from the law
Solutions 131
of conservation of mechanical energy). The frequency of the corresponding vibratory motion can
be found from the relation k/m = gb cos oc = 01%.
Therefore, having covered the distance :::0 =
(1/b) tanoc after passing the “eq}uilibrium position", the body comes to rest. At t is moment, the
restoring force "vanishes" since it is just the force
of sliding friction. As soon as the body stops, sliding friction changes the direction and becomes
static friction equal to mg sin oc. The coefficient of
friction between the body and the inclined plane
at the point where the body stops is pst == b·2:::0 =
2 tan or., i.e. is more than enough for the body to
remain at rest.
Usinglthe vibrational approach to the description of t is motion, we find that the total time of
motion of the body is equal to half the "period of
vibrations”. ‘ Therefore,
t_ _1; _ 2n _ at
2 2*% ]/gb cos oz. O
1.30. The friction Fr, (x) of the loaded sledge is
directly proportional to the length a; of the part of
the sledge stuck in the sand. We write the equation
of motion for the sledge decelerated in the sand
in the first case:
as
md-? —·mg T IJ.,
where m is the mass, a the acceleration, l the
length, and it the coefficient of friction of the
sle ge against the sand. As in the solution of
Problem 1.29, we obtained an “equation of vibrations”. Therefore, the deceleration of the sledge
stuck in the sand corresponds to the motion of /a
load on a spring (of rigidity k=(mg/li) .·. p.).__having
acquired the velocity vo in the equilibrium position.
Then the time dependence :1: (t) of the part of the
sledge stuck in the sand and its velocity v (t) can
be written as
x (t) = xo sin mot, v (t) = vo cos mot,
9*
132 Aptitude Test Problems in Physics
where
Y? `E
· <·*·»= 1/rl/·z· MIt can easily be seen that the time before the sledge
comes to rest is equal to quarter the "period of vibrations". Therefore,
t __ arm __ at/2
I" """`""` "’“ ""—"'—""` •
2‘°<> V 1/(ug)
In the second case (after the jerk}; the motion
can be regarded as if the sledge stuc in the sand
had a ve ocity v, > vo and having traversed the
distance xo were decelerated to the velocity vo
(startin5 from this moment, the second case is
observe ). The motion of the sledge after the jerk
can be represented as a part of the total vibratory
motion according to the law
a: (t) = xl sin mot, v (t) = v1 cos wot
starting from the instant tz when the velocity of
the sledge becomes equal to vo. As before, ::1 =
U1/(D0. Besides,
..m...¤._ »...2£.-m
2l *”°’ 2 2 ·
vi == $00% -‘/-2. .
The distance covered by the sledge after the
jerk is
1
x1'_x0 =·£·§·*··(%§·=·6;(”1·”o) = $0 (I/E-1)Consequently, the ratio of the braking lengths is
.€;;;&= ,/5..1,
$0
In order to determine the time of motion of the
sledge after the jerk, we must find the time of motion of the sledge from point xo to point xl by
Solutions 133
using the formula sc (t) = xl sin mot. For this
purpose, we determine tz from the formula
xo i $1 (.00t2•
Since xl = ]/2:::0, motz = rt/4. Conseqluently, t, =
sr/(4mo) = tl/2. Since ts = tl -- tz, w ere ts is the
time of motion of the sledge after the jerk, we obtain the required ratio of the braking times
z, __ 1
tl -_-l 2 •
1.31. The force of ravity mg = 60 N acting on the
load is considerably stronger than the force with
which the rope should be pulled to keep the load.
This is due to considerable friction of the rope
against the log.
At first, the friction prevents the load from
slipping under the action of the force of gravity. The
complete analysis of the distribution of friction
acting on the rope is rather complex since the
tension of the rope at points of its contact with the
log varies from F1 to mg. In turn, the force of pressure exerted by the rope on the log also varies,
being proportional at each point to the corresponding local tension of the rope. Accordingly, the
friction acting on the rope is determined just by
the force of pressure mentioned above. In order to
solve the problem, it should be noted, however,
that the total friction Fr,. (whose components are
proportional to the reaction of the log at each
point) will be proportional (avith the corresponding
progiprtionality factors) to t e tensions of the rope
at t e ends. In particular, for a certain coefficient
k, it is equal to the maximum tension: Ft,. == kmg.
This means that the ratio of the maximum tension
to the minimum tension is constant for a given
arrangement of the rope and the log: mg/ T1 =
1/(1 — k) since T1 = mg — kmg.
When we want to lift the load, the ends of the
rope as if cha-nge places. The friction is now directed against the force T2 and plays a harmful role.
The ratio of the maximum tension (which is now
134 Aptitude Test Problems in Physics
equal to T2) to the minimum tension (mg) is obviously the same as in the former case: T2/ (mg) =
1/(1 — lc) = mg/T1. Hence we obtain
2
T1
1.32. Let us see what happens when the driver turns
the front wheels of a stationary car (we shall consider only one tyre). At the initial moment, the
wheel is undeformed (from the point of view of
torsion); and the area of the tyre region in contact
with t e ground is S. By turning the steering
wheel, the driver deforms the stationary tylre until
the moment of force yizis applied to the w eel and
tending to turn it becomes larger than the maximum possible moment of static friction acting on
the tyre of contact area S. In this case, the forces
of friction are perpendicular to the contact plane
between the tyre and the ground.
Let now the motorcar move. Static frictional
forces are applied to the same region of the tyre of
area S. They almost attain the maximum values
and lie in the plane of the tyre. A small moment of
force oils applied to the wheel is sufficient to turn
the wheel since it is now counteracted by the total
moment of "oblique" forces of static friction which
is considerably smaller than for the stationary
car. In fact, in the case of the moving car, the
component of the static friction responsible for the
torque preventing the wheel from being turned is
similar to liquid friction since stagnation is not
observed for turning wheels of a'moving car. Thus,
a small torque can easily turn a moving wheel, and
the higher the velocity (the closer the static friction
to the limiting value), the more easily can the
wheel be turned.
1.33. Let us choose the reference frame as shown
—>
in Fig. 144. Suppose that vector OA is the vector
——-)—
of the initial velocity v. Then vector AB_ is the
change in velocity .during the time interval At.
Since the force acting on the body is constant.
Solutions {35
—·>· ->- _ _
vector BC equal to vector AB is the change in velocity during the next time interval At. Therefore,
1n the time interval 3At after the beginning of
.9
FS
0
B
_ 0 u A .1.*
Fig. 144
action of the force, the direction of velocity will
——> -—» -> "*
be represented by vector OD, and AB = BC : CD.
->
Let the projections of vector AB on the :1:- and
y;axes be Avx and Avy. Then we obtain two equations:
(v+Avx)”+Av; =l£-»
.2
(v+ 2Avx)’—I- (2Avy)“ = {5 ·
Since the Enal velocity satisfies the relation
v? = (v —}— 3Avx)2 -|- (3Avy)’,
using the previous equations, we obtain
vfzz .%..7. v_
1.34. Since the motion occurs in the horizontal
plane, the vertical component of the force acting
on the load is mg, and the horizontal component is
given by F2 - (mg)2 .-= mg]/4;,2 -- 1, where ¤¤ :
1.25 (Fig. 145). The horizontal acceleration of the
136 Aptitude Test Problems in Physics
load (and the carriage) is determined by this
horizontal force: ma=mg]/cx? -— 1. Consequently,
a = g)/or? — 1 == (3/4)g = 7.5 m/sz.
__ s s s
Ihcmg x 2 4
{
sl, ‘*°
”’9
Fig. 145 Fig. 146
On the first segment of the path, the carriage is
accelerated to the velocity v = atl = (3/4) gtl =
30 m/s and covers the distance sl in the forward
direction in the straight line:
z° 3
S1=-%-—=TgZ%=60 IH.
Further, it moves at a constant velocity v during
the time interval tl = 3 s and traverses the path
of length
sll = viz = 90 m.
Thus, seven seconds after the beginning of motion,
the carriage is at a distance sl -l— sz = 150 m in
front of the initial position.
On the third segment, the carriage moves round
a bend to the right. Since the velocity of the carriage moving on the rails is always directed along
the carriage, the constant (during the time interval
tl, == 25.12. s) transverse acceleration a = (3/4) g
is a centripetal acceleration, i.e. the carriage moves
in a circle at a constant velocity v: a = v2/R, the
radius of the circle being R = v2/a == 120 m. The
path traversed by the carriage in the circle is
sl, = Rep = vtll,
Solutions 137
whence the angle of rotation of the carriage about
the centre of the circle is tp == vta/R = 6.28 =
2n rad, i.e. the carriage describes a complete
circumference.
On the last segment, the carriage brakes and
comes to a halt since the acceleration along the
carriage is equal to the initial acceleration and
acts uring the same time interval. Therefore,
84 = sl = 60 m. The carriage stops at a distance
s .= 2:1 -|- s, = 210 m in front of the initial position (Fig. 146).
1.35. Let the hinge be displaced downwards by a
small distance Aa: as a result of application of
the force F, and let the rods be elongated by Al
"..’
\_r w.
7
F'
Fig. 147
(Fig. 147%. Then the rigidity k of the system of
rods can e determined rom the equation k Ax =
2k Al cos on', where 2<x’ is the angle between the
rods after the displacement. Since the displacement is small
ct' z ot, Al z Ax cos oz,
and hence k z 2k,, cosz cz.
1.36. Let us first suppose that the air drag is absent. Then the balls will meet if the vertical component of the initial velocity of the second ball is
equal to that of the first ball:
v1_= v, sin oz,
138 Aptitude Test Problems in Physics
whence sin on = vt/v2 = 10/20 ; 1/2, oc =- 30°.
Then the time of motion of the balls before collision is t =·- s/(v2 cos or.) z 0.6 s.
Since the balls are heavy, the role of the air
draglcan easily be estimated. The nature of motion
of t e hrst ball will not change significantly since
the acceleration due to the air drag is amax ==
1 m/sz even if the mass of each ball is 10 g, and
the maximum velocity of the. first ball is v1 =
10 m/s. This acceleration does not change the total
time of motion of the first ball by more than 1%.
Since the air d-rag is directed against the velocity
of the ball, we can make the balls collide by imparting the same vertical velocity component to the
second ball as that of the first ball provided that in
subsequent instants the vertical projections of the
accelerations of the balls are identical at any instant of time. For this purpose, the angle or. formed
by the velocity vector of the second ball with
the horizontal at the moment it is shot off must
be equal to 30°.
1.37. Let us write the equation of motion for the
ball at the moment when the spring is compressed by Ax:
ma = mg —- k Ax.
As long as the acceleration of the ball is positive,
its velocity increases. At the moment when the
acceleration vanishes, the velocity of the ball attains the maximum value. The spring is compressed thereby by Al such that
mg — k Al = 0,
whence
mg
AZ k .
Thus, when the velocity of the ball attains the
maximum value, the ball is at a height
mg
hZk
from the surface of the table.
Solutions 139
1.38*. It can easily he seen that the ball attains
the equilibrium position at an angle ot of dehection
of the thread from the vertical, which is determined
from the condition
tan omg _
mg
During the oscillatory motion of the ball, it will
experience the action of a constant large force
F 7- \/(mg)2 ~I— (pv)2 and a small drag force
(Fig. 148). Consequently, the motion of the ball
I
I
GI
I
I
I
Frr I
4:
I
I
r ”’9
Fig. 148
will be equivalent to a weakly attenuating motion
of a simple pendulum with a free-fall acceleration
g' given by
g. __: pe = g 1/<me>*+<nv>°
cosca mg
= 1 (g
8 + mg
140 Aptitude Test Problems in Physics
The period of small (but still damped) oscillations
of the ball can be determined from the relation
T -.= ........?..’.E_...
I/S"/I · P2/(4m’)
I/(8/Z) I/i + (IW/m8)°—·M“/(4m°)
1.39. We write the equilibrium condition for a
Small segment of the string which had the length
rs
rn
Fig. 149
Ax before suspension and was at a distance x
from the point of suspension (Fig. 149):
-% Axg+ T ($-1- Ax) = T (s),
where L is the length of the rubber string in the
unstretched state. Thus, it is clear that after the
suspension the tension will uniformly decrease
alongh the string from mg to zero.
T erefore, t e elontgations per unit length for
small equal segments o the string in the unstressed
state after the suspension will also linearly decrease
from the magimum valueto zero. For this reason,
the half-suméff the elongations of two segments of
the string symmetric about its middle will be
equal to t e elongation of the central segment which
experiences the tension mg/2. Consequently, the
Solutions 141
Q
li
i elongation Al of the string will be such as if it
were acted__upon by the force mg/2 at the point of
suspension and at the lower end, and the string
were weightless; hence
_ ms
1.40. We assume that the condition m + mz >
i- ma + m4 is satisfied, otherwise the equilibrium is
impossible. The left spring was stretched with the
ik- force TI balancing the force of gravity mzg of the
. load: T1 = mzg. The equilibrium condition for the
W load m, was
m$g+ T2 —— 1;,12011; 07
where T is the tension of the right sgring, and
" Ft is the tension of the rope passed t rough the
, pulley (see Fig. 14). This rope holds the loads of
mass ml and m2, whence
We can express the tension T2 in the following way:
T2 = (m1‘i" mz"" ms) 8After cutting the lower thread, the equations of
motion for a l the loads can be written as follows:
mm = me —l- T1 — Fam. mm = mls — T1.
in- maas Z T2 + msg '*' Fte¤• '*m4G4 Z m4g ‘ T2.
Using the expressions for the forces T1, T2, and
· Fm, obtained above, we find that
,,,.;,,:,,:0, ,4:
m4
1.41. Immediately after releasing the upper pulley,
the left load has a velocity v directed upwards,
while the right pulley remains at rest. The accelerations of the loads will be as if the free end of
the rope were fixed instead of moving at a con-
142 Aptitude Test Problems in Physics
stant velocity. They can be found from the following equations:
mal = T1 — mg, m.a2 = T2 — mg,
where m is the mass of each load, and T1 and T2
are the tensions of the ropes acting on the left and
right loads. Solving the system of equations, we
obtain al = ——(2/5)g and a2 = (1/5)g. Thus, the
acceleration of the left load is directed downwards, while that of the right load upwards. The
time of fall of the left load can be found from the
equation
0.4gt‘·*
h ·~·· vt ····*·?·— 1 0,
whence
_ 2.5v 1/`6.25v2 5h
t ` s + g“ { s °
During this time, the right load will move upwards. Consequently, the left load will be the
hrst to touch the floor.
1.42. Each time the block will move along the
inclined plane with a constant acceleration; the
magnitudes of the accelerations for the downward
an upward motion and the motion along the
horizontal guide will be respectively
al == pgcoscz —— gsin ct,
a2 = pgcoscx—|— gsin cx,
a = pg cos on
(Fig. 150). Here cz is the slope of the inclined plane
and the horizontal, and p is the coefficient of
friction. Hence we obtain
a —{-a
a:—.:...%i•
Solutions 143
The distances traversed by the block in_ uniformly varying motion at the initial velocity v
before it stops can be written in the form
U2 D2 v2
l1=T2`E’ l:7&§" LEE"
Taking into account the relations for the accelerations al, az, and a, we can find the distance I traz}®\
Ilb ’*
an
Fig. 150
versed by the block along the horizontal guide:
I: 21,1,
Zi+l¤ °
1.43. We shall write the equations of motion for
the block in terms of projections on the axis directyN
Ffr
.. mg ¤=
Fig. 151
ed downwards along the inclined. plane. For the
upward motion of the block, we take into account
all the forces acting on it: the force of gravity mg,
the normal reaction N, and friction Ft,. (Fig. 151),
144 Aptitude Test Problems in Physics
and obtain the following equation:
mg sin oc -l— p.mg cos on = mal.
The corresponding equation for the downward motion is
mg sin ct — p,mg cos cz = mag.
Let the distance traversed by the block in the
upward and downward motion- be s. Then the time
of ascent tl and descent tz can be determined from
the equations
22
8 ,7; , 8; •
By hypothesis, 2tl = ta, whence 4a2 = al. Consequently,
gsinct -|- pgcoscz = 4 (gsin on — pgcos cx),
and finally
p, =—- 0.6 tan cc.
L44. If the lower ball is very light, it starts climbing the support. We shall find its minimum mass
_ “¤
Jr/?·2a· *
N. » is i
Z ”'¢9
cz as
(rQ1
xl m2g
Fig. 152
mz for which it has not yet started climbing, but
has stopped pressing against the right inclined
plane. Since the support is weightless, the horizontal components of the forces of pressure (equal
in magnitude to the normal reactions) exerted by
the ba ls on the support must be equal (Fig. 152);
&nlutions 145
ihtherwise, the "support" would acquire an inhnitely
Eflarge acceleration:
sin cc = N2 sin ot, N2 = N2.
§§Moreover, since the lower ball does not ascend,
tjhe normal components of the accelerations of the
Etballs relative to the right inclined plane must be
abqual (there is no relative displacement in this
=sdirection). Figure 152 shows that the angle between
the direction of the normal reaction N2 of the suport and the right inclined plane is rt/2 ——- 2a., and
gence the latter condition can be written in the
form
m2gcoscz—N1 _ m2gcosoc—N2 cos 2ot
4 mi * mz ,
whence m2 = ml cos 20c. Thus, the lower ball will
"climb" up if the following condition is satisfied:
1.45. As long as`the cylinder is in contact with
the suprilorts, the axis of the cylinder will be exactly at t e. midpoint between the supports. Consequently, the horizontal component of the cylinder
velocity is v/2. Since all points of the cylinder axis
move in a circle with the centre at point A, the
total velocity u of each point on the axis is perpendicular to the radius OA =r at any instant of time.
Consequently, all points of the axis move with
a centripetal acceleration dc = u2/r.
We shall write the equation of motion for point
0 in terms of projections on the "centripetal" axis:
z
mgcosoc——·N=ma2 ==—’€f——- , (1)
where N is the normal reaction of the stationary
support. The condition that the separation between
the supports is rg 2 implies that the normal reaction of the mova le support gives no contribution
to the projections on the "centripetal" axis. Accordmg to Newton’s third law, the cylinder exerts
10-0771
146 Aptitude Test Problems in Physics
the force of the same magnitude on the stationary
support. From Eq. (1), we obtain
N = mg cos on —— .
At the moment when the distance between points
A [arid B of the supports (see Fig. 18) is AB =
ry 2, we have
eos cx := =—L· .
Zr l/2
The horizontal component of the jelocity of point O
is u cos o¢=v/2, whenceu= v]/2. Thus, for AB =
1*]/ 2, the force of normal pressure exerted by the
cylinder is
__ mg _mv2
N_·· *"'E';_" •
For the cylinder to remain in contact__with the supports until AB becomes equal to r]/ 2, the condition g/1/2 > v2/(2r) must be satisfied, i.e. v.<
V srl/Z
1.46. The cylinder is acted upon by the force of
gravity mlg, the normal reaction N1 of the left
$2
\W
cz ”'19 A W
$1
Fig. 153
inclined plane, and the normal reaction N 3 of the
wedge (force N 3 has the horizontal direction). We
shall write the equation of motion of the cylinder
in terms of projections on the xl-axis directed
along the left inclined plane (Fig. 153):
mlal = mlg sin ot — N3 cos cx, (1)
Solutions 147
where al is the projection of the acceleration of the
cylinder on the xl-axis. °
The wedge is acted upon by the force of gravity
mzg, the normal reaction N 2 of the right inclined
plane, and the normal reaction of the cylinder,
which, according to Newton’s third law, is equal
to —N3. We shall write the equation of motion of
the wedge in terms of projections on the :1;,,-axis
directed along the right inclined plane:
ma, = —m2g sin cx + N 8 cos ct. (2)
Du1·ing its motion, the wedge is in contact
with the cylinder. Therefore, if the displacement
of the wedgejalong the sc,-axis is Ax, the centre of
the cylinder (together with the vertical face of the
wedge) will be displaced along the horizontal by
Ax cos oa. The centre of the cylinder will be thereby
displaced along the left inclined plane (xl-axis) by
Ax. This means that in the process of motion of
the wedge and the cylinder, the relation
**1 = az (3)
is satisfied.
Solving Eqs. (1)-(3) simultaneously, we determine the force of normal pressure N =N 3 exerted by
the wedge on the cylinder:
2m m
N = ——-Li tan oc.
’ m1+m2
L47. As long as the load touches the body, the velocity of the latter is equal to the horizontal component of the velocity of the load, and the acceleration of the body is equal to the horizontal component of the acceleration of the load.
Let a be the total acceleration of the load.
Then we can write a = at + ac, where ac is the
centripetal acceleration of the load moving in the
circle of radius l, i.e. dc = v2/l, where v is the
velocity of the load (Fig. 154). The horizontal
component of the acceleration is
2
ah: at sin ot——%— cos oc.
10*
148 Aptitude Test Problems in Physics
The body also has the same acceleration. We can
write the equation of motion for the body:
2
N: Mah: Mat sin oc-M gi- cos oc,
where N is the force of normal pressure exerted
ac.
it ·
l. ue M
A II.
./x//
Fig. 154
by the load on the body. At the moment of separation of the load, N = 0 and
. v2
dt SHI @1: T COS G.
The acceleration component at at the moment of
separation of the load is only due to the force of
gravity:
at = g COS OL.
Thus, the velocity of the load at the moment of
separation is
v: }/gl sin oc,
and the velocity of the body at the same moment is
u:vsin oc: since 1/glsinoz
According to the energy conservation law, we have
2
mgl: mgl sin oc—{—-%lL——§— Mvz sinz -—g— _
Substituting the obtained exgression for v at the
moment of separation and t e value of sin oc_=
Solutions 149
sin at/6 = 1/2 into this equation, we obtain the
ratio:
M 2——— 3 sin on
•—-·~ i 7;.* 4.
m sm on
The velocity of the body at the moment of separation is
. 1 1/ gl
u - v sm cz .. 2 2 ,
1.48. The rod is under the action of three forces:
the tension T of the string, the force of gravity mg,
and the reaction of the wall R = N + Fr,. (N is
the normal reaction of the wall, and Ft,. is friction,
Ft,. Q pN). When the rod is in equilibrium, the
sum of the moments of these forces about any point
is zero. For this condition to be satisfied, the line
iof action of the force R must pass through the
point of intersection of the lines of action of T
and mg (the moments of the forces T and mg
gabout this point are zero).
;,, Depending on the relation between the angles oz.
Qand B, the point of intersection of the lines of acition of T and mg may lie (1) above the perpentdicular AMO to the wall (point M1 in Fig. 155);
Q2) below this perpendicular (point M2); (3) on the
§erpendicu1ar (point M 0). Accordingly, the friction
be either directed upwards alongA C (Fm), or downy ards along AC (Fh.2), or is equal to zero. Let us
nsider each case separately.
(1) The equilibrium conditions for the rod are
cos on —{- Ff,.1 —— mg = 0, N — T sin oz. == 0 (1)
i1`!._
he sums of the projections of all the forces on the
and y-axes respectively must be zero), and the
yi ments of forces about point A must also be zero:
d mgl . Tl .
4,: : .1 = Tdz, or -5- sm B: T3- sin (oc+B), (2)
150 Aptitude Test Problems in Physics
where dl and d2 are the arms of the forces mg and
T respectively. From Eqs. (1) and (2), we obtain
H > Fm}; Si¤(¤¤+B>__ 1
1/ N 3 sinocsinb tance
___ _1__ ( 2 ___ 1 )
" 3 tanfi tanoc °
This case is realized when 2 tan on > tan B.
T\
a
*6
‘
.~ 0
gf "¤
X"
M2
Fig. 155
(2)) After writing the equilibrium conditions,
we o tain
112
*122-3- ( tancc _ tanfi
Thisbcase corresponds to the condition 2 tan on <
tan .
(3) In this case, the rod is in equilibrium for
any value of pa: 2 tan or. == tanrb.
Solutions 151
Thus, for an arbitrary relation betw&en the
angles cc and 6, the rod is in equilibrium
112
p' >-3-1 tanoc — tanfl
1.49*. Let us analyze the motion of the smaller
disc immediately after it comes in contact with the
larger disc. .
We choose two equal small regions of the smaller disc lying on the same diameter symmetrically
about the centre O' of this disc. In F1g. 156,
Ffr2.
\Ffrl
V1
Ar
\ uy
\\ r h
x\ O W [2
sz / . "
A"
0A
Fig. .56
poirts A2 and A 2 are the centres of mass of these region. At the moment of contact (when the smaller
discis still at rest), the velocities v1 and U2 of the
poiits of the larger disc which are in contact with
poiats A2 and A2 of the smaller disc are directed as
shovn in Fig. 156 (v2 = $2-0Al and v2 = S2·0A2).
The forces of friction F2,.2 and Ff,.2 exerted by the
larter disc on the centres of mass A2 and A2 of
the selected regions of the smaller disc will obviouly be directed at the moment of contact along
the velocities v1 and U2 (Frm = F222). Since the arm
l of the force F;2.1 about the axis of the smaller
disc is smaller than the arm 12 of the force FH2
(ee Fig. 156), the total torque of the couple Fm
152 Aptitude Test Problems in Physics
and Fha will rotate the smaller disc in the direction of rotation of the larger disc.
Having considered similar pairs of regions of
the smaller disc, we arrive at the conclusion that
immediately after coming in contact, the smaller
disc will he rotated in the direction of rotation of
the larger disc.
Let the angular velocity of the smaller disc at
a certain moment of time become oa. The velociFfr2
Flrl 6.
x` --05 B2
’ \ U2
'
B1 E12
Ur
.- s'
. A1 L- U2
\
v' \ ·
7 \\ 0 (‘}
Q\
.» °’
042
1
Fig. 157
ties of the regions with the centres of mas at
points A1 and A, will be vi = vg = mr, vhere
r = 0’A1 = O'A, (Fig. 157). The forces of friction
Flin and Firz acting on these regions will be directed along vectors v1———v§ (the relative veloity
of the point of the larger disc touchingl pointA1)
and v2 -— vg (the relative velocity of t e poin of
the larger disc touching point A,). Obviously, the
torque of the couple Flin and F;1.2 will accelente
the smaller disc (i.e. the angular velocity of he
disc will vari) if vj = vg < BIB2/2 = Qr (ee
Fig. 157; for t e sake of convenience, the vecttrs
“pertaining" to point A2 are translated to point A),
_ Thus, as long as co < S2, there exists a nonzeo
frnctional torque which sets the smaller disc inc
Solutions 153
rotation. When co = S2, the relative velocities of
the regions with the centres of mass at points A1
and A2,are perpendicular to the segment OO' (directed along the segment AIC in Fig. 157), and the
frictional torque about the axis of the smaller disc
is zero. Consequently, the smaller disc will rotate
at the steady·state angular velocity $2.
For co = Q, all the forces of friction acting on
similar pairs of regions of the smaller disc will be
equal in magnitude and have the same direction,
viz. perpendicular to the segment OO'. According
to Newton’s third law, the resultant of all the
forces of friction acting on the larger disc will be
applied at the point of the larger disc touching the
centre O' of the smaller disc and will be equal to
umg. In order to balance the decelerating torque of
this force, the moment of force
chi = pmgd
must be applied to the axis of the larger disc.
1.50. After the translatory motion of the system
has been established, the ratio of the forces of friction Fm and Fm, acting on the first and second
rods will be equal to the ratio of the forces of pressure of the corresponding regions: Fin/F", =
N1/N2. Since each force .of pressure is proportional
to the mass (N1 = mlg and N, = mzg), the ratio
of the forces of friction can be written in the form
Ffrl ____ ml .1
Ftrs mz O ( )
On the other hand, from the equality of the moments of these forces about the vertex of the right
angle (Fig. 158, top view), we obtain
IF", cos q> = IFH2 sin qa, (2)
where Z is the distance from the vertex to the centres of mass of the rods. From Eqs. (1) and (2), we
obtain tan q> = ml/mz, where q> = cx. — n/2. Consequently, on = rc/2 1; arctan (ml/mz).
1.51. If the foot of t e footba l player moves at a
Velwity u at the moment of kick, the velocnty of
154 Aptitude Test Problems in Physics
the ball is v + u (the axis of motion is directed
along the motion of the ball) in the reference frame
fixed to the foot of the player. After the perfectly elastic impact, the velocity of the ball in the
as
Ffrl p
I, "/ Y .
Ffr2
Fig. 158
same reference frame will be —(v + u), and its velocity relative to the ground will be ——(v —l— u) —
u. If the ball comes to a halt after the impact,
v -I— 2u = O, where u = -—v/2 = -5 m/S. The minus sign indicates that the foot of the sportsman
must move in the same direction as that of the
ball before the impact.
1.52. Since in accordance with the momentum
conservation law, the vertical component of the
velocity of the body-bullet system decreases after
the bullet has hit the body, the time of fall of the
body to the ground will increase.
In order to determine this time, we shall find
the time tl of fall of the body before the bullet
h1_ts it and the time tz of the motion of the body
with the bullet. Let to be the time of free fall of
the body from the height h. Then the time in which
the body falls without a bullet is tl = ]/fh/g =
t /1/2. At the moment the bullet of mass M hits
the body of mass m, the momentum of the body is
directed vertically downwards and is
mgtc
mv`: ······:.—· .
V2
Solutions 155
The horizontally flying bullet hitting the body
will not change the vertical component of the momentum of the formed system, and hence the vertical component of the velocity of the body—bullet
system will be
u - __...”‘ ,,.. ....”.';.. ..*3;..
_ m—|—M — m—|—M g l/§ ‘
The time tz required for the body—bullet system to traverse the remaining half the distance
can be determined from the equation
2
gi: ***52+ ·%g· .
This gives
t to I/m”+(m+M)“—-m
1/2 m—l—M ‘
Thus, the total time of fall of the body to the
ground (M > m) will be
to I/m”+<m+M)“-i—M r·
t= *···:· % to ll 2.
1/ 2 m+
1.53. In order to solve the problem, we shall use
the momentum conservation law for the system.
We choose the coordinate system as shown in
Fig. 159: the a:-axis is directed along the velocity
v of the body of mass ml, and the y—axis is directed
along the velocity v of the body of mass m,.
After the collision, the bodies will stick together
and fly at a velocity u. Therefore,
mm = (ml + mz) use mm = (ml + mz) uyThe kinetic energy of the system before the collision
was
1 ____ ntl')? m2vg
Wk `° 2 + 2 ·
,156 Aptitude Test Problems in Physics
—` The kinetic energy of the system after the collision
(sticking together) of the bodies will become
22
w~ = .E!..`lLL"2. 2 2 ;.=_”‘.1.”¢_`l"”..2"2._
" 2 (u°°+u") 2 <mi+m2>
Thus, the amount of heat liberated as a result of
collision will be
=W’—W" :-2]}-T2>—— ” 2 zii.3 J.
Q K k 2<mi+m2) ("*+"¤)
HI
•€· .
l’"2”2
Fig. 159
1.54. Since there is no friction, external forces do
not act on the system under consideration in the
Fig. 160
horizontal direction (Fig. 160). In order to determine the velocity v of the left wedge and the velocity u of the washer immediately after the descent,
Solutions 157
we can use the energy and momentum conservation
laws:
z2
A-l—q———{—in—ui—=mgh, Mvzmu.
22
Since at the moment of maximum ascent lrmax
of the washer along the right wedge, the veloc1t1es
of the washer and the wedge wi 1 he _equal_, the
momentum conservation law can be written in the
form
mu = (M -I— m) V,
where V is the total velocity of the washer and the
right wedge. Let us also use the energy conservation law:
2M
-i"-§—-—{'—€”— v2+mghm..x.
The joint solution of the last two equations leads
to the expression for the maximum he1ght hmax
of the ascent of the washer along the right wedge:
Mz
h = h -——————— .
max ]n)2
1.55. The block will touch the wall untilthe washer comes to the lowest position. By this nnstant of
*2%
Fig. 161
time, the washer has acquired the velocity v which
can be determined from the energy conservation
law: v2 = 2gr. During the subsequent motion of
the system, the washer will "c1imb" the r1ght-hand
side of the block, accelerating it all _the t1me in
the rightward direction (Fig. 161) unt1l the veloc-
158 Aptitude Test Problems in Physics
ities of the washer and the block become equal.
Then the washer will slide down the block, the
block being accelerated until the washer Easses
through the lowest position. Thus, the bloc will
have the maximum velocity at the instants at
which the washer passes through the lowest position during its backward motion relative to the
block.
In order to calculate the maximum velocity of
the block, we shall write the momentum conservation law for the instant at which the block is separated from the wall:
mz V?-é?= miv1+ mzvz.
and the energy conservation law for the instants at
which the washer passes through the lowest position:
_ mw? mv?
mzgr- 2 -|- 2 ,
This system of equations has two solutions:
(1) vi=0. vz= 1/EE;.
2m — m ——m ——-2 == .. 2 , :.;.2.. 2 _
()v1 m1+mz V gr U2 m1+mz 1/ gr
Solution (1) corresponds to the instants at which
the washer moves and the block is at rest. We are
interested in solution (2) corresponding to the instants when the block has the maximum velocity:
v _ 2mz
1max m1+m2 .
1.56. Let us go over to a reference frame fixed to
the box. Since the impacts of the washer against
the box are perfectly elastic, the velocity of the
washer relative to the box will periodically reverse
its direction, its magnitude remaining equal to v.
It can easily be seen that the motion of the washer
will be repeated with fperiod 2At, where At =
(D — 2r)/ v is the time 0 flight of the washer be-
Solutions 159
(tween two successive collisions with the box (every
time the centre of the washer covers a distance
(D — 2r at a velocity v).
` Returning to the reference frame fixed to the
ground, we can plot the time dependence vwash (t)
of the velocity of the centre of the washer. Knowing the velocity graph vwash (t), we can easily
plot the time dependence of the displacement
xwash (t) of the centre of the washer (Fig. 162).
Uwi1Sh
20 ·
. p-gp p-gp
uv
1
z“
—Z`w¤Sh I
U-2r· ·
t7="D`" " "
ta vt, 21*, 52.*, 4t, 52*; T
Fig._162
1.57. Thelforces acting on the]hoop-washer system
are the force of gravity and the normal reaction of
the lplane. These forces are directed along the vertica . Consequently, the centre of mass of the
system does not move in the horizontal direction.
Since there isQno friction between the hoop and
the plane, the motion of the,hoop is translatory.
According to the momentum _conservation law, at
any instant of time we have
Mu —I- mvx = 0, (1)
where u and vx are the horizontal comgonents of
the velocities of the centre of the hoop an the washer. Since vx {periodically changes its sign, u also
changes sign ‘synchronously". The general nature
160 Aptitude Test Problems in Physics
of motion of the hoop is as follows: the centre of
the hoop moves to the right when the washer is onsegments BC and BE, and to the left when the
washer is on segments CD and DE (Fig. 163).
.(.
\ ”’”‘
A\
//x
Fig. 163
The velocities v of the washer and u of the hoop
are connected through the energy conservation law:
a2
mzr(1+<><>¤ <»>>=-’i‘§——+-"-Qi-. <2>
The motion of the washer relative to a stationary observer can be represented at any instant as
the superposition of two motions: the motion relative to the centre of the hoop at a velocity vt directed along the tangent to the hoop, and the motion together with the hoop at its velocity u having
the horizontal direction (Fig. 164). The figure shows
that
Dy —tan 3
,,x+,,y ‘P· ( )
Solving Eqs. (1)-(3) together, we determine the
velocity of the centre of the hoop at the instant
when t e radius vector of the point of location of
the washer forms an angle cp with the vertical
_ l , 2gr (1 —}-cos q>)
“"’”°°S ‘*’ 1/ (M-pm) (M-.·-msamp) ·
Solutions 161
1.58. At the moment of snaplling of too right
string, the rod is acted upon bY the tooslon T Of
.9
ge
n //S°
U Xi':
\\\ A lx
J? ____ ”#
Fig. 164
the left string and the forces N1 ang N2 oogggil
pressure of the loads of mass gn; an m? ( lg') thé
Since the rod is weightless (1ts mass lo zoro ·
4/ ’// f/.’/’./ I’/,",’.’/’{’.’/’// /_.// ,
N;
T N2
w, ’”29
""1!l
Fig. 165
equations of its translatory and 1‘0m1`Y motions will
have the form
.-T + N, - N, = 0, _N1l ; 2N¤'»
1y·0771
162 Aptitude Test Problems in Physics
The second equation (the condition of equality to
zero of the sum of all moments of force about point
0) implies that
N1 i ZN2.
Combining these conditions, we get (see Fig. 165)
T—_=N1‘*N2;N2•
At the moment of snapping of the right string,
the accelerations of the oads of mass ml and ml
will be vertical (point O is stationary, and the rod
is inextensible) and connected through the relation
al = 2al. (3)
Let us write the equations of motion for the loads
at this instant:
mig ‘“‘ Ni = mlalv mag 'l‘ Ni = mzam
where N { and N g are the normal reactions of the
rod on the loads of mass ml and ml. Since N; = Nl
and Ng = N2, we have
mlg — 2N2 == mlal, mag + N, = Zmlal.
Hence we can find N , and consequently (see
Eq. (2)) the tension of the string
... _ ._{’h.’i‘.a...
T- Na — ”'1+ 4me g.
1.59. Let the ring move down from point A by a
distance Ax during a small time interval At elapsed
A
4 Ae A
Fig. 166
after the beginning of motion of the system
and acquire a velocity v (Fig`. 166). The velocity of
translatory motion of the oops at this moment
Solutions {63
must be equal to u = v tan on (At is so small that
the angle on practically remains unchanged). Consequently, the linear velocity of all points of the
hoopls must have the same magnitude. According
to t e energy conservation law, we have
mg Ax=2Mu¤+-'Z;3-=2Mv¤ta¤¤a+L"2'?; ,
where M uz is the kinetic energy of each hoop at a
given instant. From this equality, we obtain
..2i......_...c~._...... ..__J..__
2A:c _ 4M tan“ot—|-m g- l—|-4(M/m)tan°ot g'
As Ax —» 0, we can assume that v2 = 2a Ax, where
a is the acceleration of the ring at the initial instant of time. Consequently,
1
a " 1—|—4(M/m) tan2oc g'
1.60. Let the rope move over a distance Al during
a small time interval At after the beginning of motion and acquire a velocity v. Since At is small,
we can assume that
v2 =—- 2a Al, (1)
where a is the acceleration of all points of the rope
at the initial instant.
From the energy conservation law (friction is
absent), it follows that
Mvz
T=AWp,
where M is the mass of the rope, and AWD is the
change in the potential energy of the rope during
the time interval At. Obviously, AWD corresponds
to the redistribution of the mass of the rope, as a
result of which a piece of the rope of length Al
"passes" from point A to point B (see Fig. 31).
Therefore,
M
awp: (T-) gh Az. (3)
11•
164 Aptitude Test Problems in Physics
From Eqs. (1)-(3), we find the condition of motion
for the rope at the initial instant of time:
gh
aI_
1.61. It is clear that at the moment of impact, only
the extreme blocks come in contact with the washer. The force acting on each such block is perpendicular to the contact surface between the washer
and a block and passes through its centre (the
diameter of the washer is equal to the edge of the
block!).— Therefore, the middle block remains at
rest as a result of the impact. For the extreme
blocks and the washer, we can write the conservation law for the momentum in the direction of the
velocity v of the washer:
mv; ..|-my'.
Here m is the mass of each block and the washer,
v' is the velocity of the washer after the impact,
and u is the velocity of each extreme block. The
energy conservation law implies that
v2 = 2u2 -I- v’2.
As a result, we find that u = v]/2 and v' = O.
Consequently, the velocities of the extreme blocks
after the imgact form the angles of 45° with the
velocity v, t e washer stops, and the middle_block
remains at rest.
1.62. In this case, the momentum conservation law
can he applied in a peculiar form. As a result of
explosion, the momentum component of the ball
along the pipe remains equal to zero since there is
no friction, and the reaction forces are directed at
right angles to the velocities of the fragments.
Inelastic collisions do not change the longitudinal
momentum component either. Consequently, the
final velocity of the body formed after all collisions
is zero.
1.63. For the liberated amount of heat to be gmaximum, the following conditions must be satisfied:
Solutions 165
(1) the potential energy of the bodies must be
maximum at the initial moment;
(2) the bodies must collide simultaneously at the
lowest point of the cu ;
(3) the velocity of) the bodies must be zero
immediately after the collision.
If these conditions are satisfied, the whole of the
initial potential energy of the bodies will be
transformed into heat. Consequently, at the initial instant the bodies must be arranged on the
brim ofthe cup at a height r above the lowest
point. The arrangement of the bodies must be such
that their total momentum before the collision is
zero (in this case, the body formed as a result of
collision from the bodies stuck together will remain at rest at the bottom of the cup). Since the
values of the momenta of the bodies at any instant
are to one another as 3:4:5, the arrangement of the
m
,. al
/·
I
{p
———-——- -m
/j§§_"}€’ 0 J
gm
Fig. 167
bodies at the initial instant must be as in Fig. 167
(top view). After the bodies are left to themselves,
the amount of heat Q liberated in the system is
maximum and equal to 4mgr.
1.64. Let the proton be initially at rest relative to
a stationarf reference frame, and let the ot-particle
have a ve ocity vo. The process of their elastic
collision is described by the momentum conservation law
4mvo = mvl —I- 4mv,
166 Aptitude Test Problems in Physics
and by the energy conservation law
4mv§ ___ mv? + 4mv§
2 _` 2 2 ·’
where U1 and v2 are the velocities of the proton and
the oc-{article in the stationary reference frame
after t e collision, and m and 4m are the masses
of the proton and the oc—particle respectively.
Let us consider the collision of these particles
i11 the centre-of-mass system, i.e. in an inertial
reference frame moving relative to the stationary
reference frame at a velocity
v,_ 4mvO _ ii- U
— m—}—4m — 5 °
(the numerator of the 6rst fraction contains the
total momentum of the system, and the denominator contains its total mass). Figure 168 shows the
•—-—---———-- -~ --)Uo
ze'- -------?
,0U
T............. B ix
/
/
//
A"A,,..U.%..`%&] 2}
Fig. 168
velocity in and the velocities of the <x—particle
—>(vector OB) and the proton (vector OA) in the centre-of-mass system before the collision: OB =
(1/5)v0 and OA = (4/5)v0. According to the momentum conservation law, after the collision, the
—> —>velocity vectors OB and OA of the on-plarticle and
the proton must lie on the same straig, t line, and
the relation OB':OA' -·= 1:4 (see Fig. 168) must
be satisfied. According to the energy conservation
law, OB' = OB and OA' =; OA (prove this!).
Solutions 167
In the stationary reference frame, the velocities of the oz—particle and the proton are represent·—-P —>·
ed-)- in thgphgure by vectors OC, = OB' —|— v' and
OC1 = OA' + v'.
In order to solve the problem, we must detfmine tbe maximum possible length of vector OC1,
i.e. in the isosceles triangle OA’C1, we must determine the maximum possible length of the base
for constant values of the lateral sides. Obviously,
—>the maximum magnitude of OC, is equal to
2 >< (4/5) vo = 1.6v,,. This situation corresponds
to a central collision.
1.65. The tyres of a motorcar leave a trace in the
sand. The higher the pressure on the sand, the
deeper the trace, and the higher the probability
that the car gets stuck. If the tyres are deflated
considerably, the area of contact between the tyres
and the sand increases. In this case, the pressure
on the sand decreases, and the track becomes more
shallow.
1.66. At any instant of time, the complex motion
of the body in the pipe can be represented as the
superposition of two independent motions: the
motion along the axis of the piipe and the motion
in the circle in a Elane perpen icular to the tpipe
axis (Fig. 169). T e s¢=§1aration of the body rom
the pipe surface will a ect only the latter motion
(the body will not move in a circle). Therefore,
we shall consider only this motion.
The body moving in a circle ex eriences the
action of the normal reaction N of tgeupipe walls
(vector N lies in the plane perpendic ar to the
pipe axis) and the "iorce of (gravity" mg' =
mg cos oa. We shall write the con ition of motion
for the body fin a circle:
mv?
mg' cosB+N=—;— , (1)
where B is the angle formed by the radius vector
of the point of location of the body at a given instant and the "vertical" y':(Fig. 170). For the body
168 Aptitude Test Problems in Physics
to remain in contact with the surface of the pipe
the condition N = mv?/r ·-— mg' cos B > 0 must
be fulfilled, whence
v2 ; g’r cos B. (2)
The relation between the velocity v at which
the body moves in a circle at a given instant and
‘Ogc`<i_5_ B
t Us \
A\T
on sm ya
g2g coscz
9 _ Q _______ > U/fig,
/3 y'
Fig. 169 Fig. 170
the initial velocity v— can be obtained from the
energy conservation law: for any value of the
angle B, the following relation must hold:
....."’_,;’” +mg·mSp=.....m<"¤;i“ W +mg·r,
whence
v2 = v§ sin’ q> -l- 2g'r — 2g’r cos B. (3)
Substituting Eq. (3) into (2), we obtain the
values of vo for which the bo y remains in contact
with the pipe:
3g’r cos B 2g' r
"i > Ti&a?F—‘;,im;·
Since this condition must be satisfied for any value
of BQ [0, 2n], we finally obtain
g’r _ gr cosor.
V3 2 Sim ¢v " ¤i¤*..q> °
Solutions 169
1.67*. Let us suppose that at a certain instant, the
wheel is in one of the positions such that its centre of mass is above a rod, and its velocity is v.
At the moment of the impact against the next rod
(Fig. 171), the centre of mass of the wheel has a
ss"
"
J
Fig. 171
certain velocity v' perpendicular to the line connecting it to the previous rod. This velocity can be
obtained from the energy conservation law:
mvz mv' 2
m8h"l"*2_ —- T .
Here h = r — ]/r2—l2/4 z lz/(8r). Therefore,
,_ gl“
v — v 4TUa •
By hypothesis (the motion is without jumps), the
impact of the whe el against the rod is perfectly
ine astic. This means that during the impact,
the projection of the momentum of the wheel on
the straight line connecting the centre of the
wheel to t e rod vanishes. Thus, during each collision, the energy
__ m(v' sin 0t)2
170 Aptitude Test Problems in Physics
where sin ot z l/r, is lost (converted into heat).
For the velocity v to remain constant, the work
done by the tension T of the rope over the path l
must compensate for this energy loss. Therefore,
__ mv? glz lz
Tl- 2 (1+ 4rv2 )7·T’
whence
_ mvzl gl2 ) ~ mvzl
T — 2rz (1+ 4rv2 N 2r2 °
1.68. Since the wheels move without slipping, the
axle of the coupled wheels rotates about point 0
while passing through the boundary between the
F
\ . ~` `\\\\~0 _ _.................r -..@;>.
Fig. 172
planes (Fig. 172). At the moment of separation,
the force of pressure of the coupled whee s on the
plane and the force of friction are equal to zero,
and hence the angle B at which the separation
takes place can be found from the condition
a
mg cos B: L-nil ,
From the energy conservation law, we obtain
z2
ng-; Egl.-mgr (1—-cos B).
Solutions 171
No separation occurs if the angle B determined
from these equations is not smaller than oc, and
hence
cos B { cos oz.
Therefore, we find that the condition
v Q ]/gr (3 cos 0c—2)
is a condition of crossing the boundary between
the planes by the wheels without separation. If
3 cos on ——- 2 < O, i.e. on > arccos (2/3), the separation will take place at any velocity v.
1.69. At the initial moment, the potential energy
of the system is the sum of the potential energy
mg (r -I— h) of the rim and the potential energy
pgh2/ (2 sin ot) of the part of the ribbon lying on t e
inclined plane'. The total energy of the system in
the final state will also be a purely potential
energy equal to the initial energy in view of the
absence of friction. The final energy is the sum of
the energy mgr of the rim and the energy of the
ribbon wound on it. The centre of mass of the latter
will be assumed to coincide with the centre of mass
of the rim. This assumption is justified if the
length of the wound ribbon is much larger than the
length of the circumference of the rim. Then the
potential energy of the wound ribbon is
P ( + S) g*~
the length of the ribbon being h/sin on —|— s, wher e s
is the required distance traversed by the rim from
the foot of the inclined plane to the point at which
it comes to rest.
From the energy conservation law, we obtain
h2 __ h
m€("+h)+P8’ ·§;TH·&··=m§· +P€T (qa? + 8) ,
s
whence
· __ mg—}—p (Iz./sin cx) (r--h/2)
P" °
172 Aptitude Test Problems in Physics
1.70. The steady-state motion of the system in air
will be the falling of the balls along the vertical at
a constant velocity. The air drag F acting on the
lower Sheavier) and the upper ball is t e same
since t e balls have the same velocity and size.
Therefore, the equations of motion for the balls
can be written in the form
mlg-—T—F=0, m,g—|—T——F=O.
Solving this system of equations, we obtain the
tension of the thread:
_ <mi—mz)g
T .. 2 _
1.71*. At each instant of time, the instantaneous
axis of rotation of the ball passes through the point
of contact between the thread and the cylinder.
This means that the tension of the thread is perpendicular to the velocity of the ball, and hence
it does no work. Therefore, the kinetic energy of
the ball does not change, and the magnitude of its
velocity remains equal to v.
In order to determine the dependence Z (t), we
mentally divide the segment of the thread unwound by the instant tintoa very large number N
of small equal pieces of len th Al = l/N each.
Let the time during which ilglé nth piece is unwound be Atn. During this time, the end of the
thread has been disp aced by a distance v Atn, and
the thread has turned through an angle Acpn =
v Atn/(n Al) (Fig. 173). The radius drawn to the
point of contact between the thread and the
cylinder has turned through the same angle, i.e.
Al
A€Pn=A€P =·*;· »
whence
2
Atl`: n(Al) •
vr
Solutions 173
Then
t=At1-|—At2+ . ..+A¢N=i-gi-igi
2 (Al)? N (Al)? ___ (Al)2 N (N+ 1)
+-2;*+ +—.;—·";;‘ *7*Since N is large, we have
__ N2 _- [2 _' ....
t—-····—%;··—·---··2;·_··, l— 1/-2UTt.
04t,,
°? F
"*’"
A
Fig. 173
1.72. During the time _T, the distance covered by
the blue ball is co (1/ 1/3) T = 2rcl/ 1/3 (Fig. 174),
where to = 2:1;/ T is the rotational frequency. During the same time, the centre of mass of the green
aQ
° “Z;27Jz‘ ‘ *3*
I
I
g
•:•-”!4`-{57 _ I -···l·
99W
Fig. 174
and the white ball will be displaced by a distance
to (l/2 ]/3) T=:rml/ ]/3. The rod connecting the green
174 Aptitude Test Problems in Physics
and the white ball will simultaneously turn
through an angle 2rt since the period of revolution
of the balls around their centre of mass coincides
with the period T. Therefore, the required distance is
""`""`"`T"'““`“T2.'
Lzzl/-ii-+(¤i/a-|—}-)
42
or for another arrangement of the balls (the white
and the green ball change places in the figure),.
3 I T 1. 2
L:] l/"`4`+l"l/'5"Y)
1.73. The centre of mass of the system consisting
of the blocks and the thread is acted upon in the
horizontal direction only by the force exerted by
the Eulley. Obviously, the horizontal component
of t is force, equal __to T (1 — cos cp), where T
is the_tension of the thread, is always directed to
_T
,, Y
so T
Fig. 175
the right (Fig. 175). Sinc_e at the initial moment the
centre of mass is at rest above the pulley, during
motion it will be displaced along the horizontal to
the 1·ight. Hence it ollows that the left block
reaches the pulley before the right block strikes the
table since otherwise the cent1·e of mass would be
to the left of the pulley at the moment of impact.
1.74. According to the initial conditions (the left
load is at rest, and the right load acquires the velocity v), the left load will move in a straight line,
while the right load will oscillate in addition to the
Solutions 175
motion in a straight line. At a certain instant, the
left load is acted upon along the vertical by a force
mg —— T, and the right load by a force mg —
Tcos cp Fig. 176, the vertical axis is directed
downwards). Here T is the tension of the thread.
•_ I if
{ oi
/
I
T L T, T2
'P dup
T
Tk
[lj *1
'"9
mg mfg ’”z9
Fig. 176 Fig. 177
Hence it follows that the difference in the vertical
components of the accelerations of the right (al)
and left (a,) loads, given by
TT
a1"'a2 = (8’*·j,·z·00S€P)··(g·"·”·,f)
T
= 7,,- (1-—¢<>S <1>>,
is always nonnegative. Since at the initial moment
the relative distance and the relative vertical velocity of the loads are equal to zero, the difference
in the ordinates of the right and left loads will
increase with time, i.e. at any instant the right
load is lower than the left one.
1.75. Let the right and loft threads be 'deflected
respectively through angles B and ot from the vertical (Fig. 177). For the rod to remain in the vertical position, the following condition must be
satisfied:
T1 sin ot = T, sin 6, (1)
176 Aptitude Test Problems in Physics
where T1 and T, are the tensions of the relevant
threads.
Let us write the equations of motion for the
two bodies in the vertical and horizontal directions:
T1 sin ot = mlonzll sin oc, T1 cos on = mlg,
T, sin [5 = mzmzlz sin 6, T2 cos 6 = mzg.
Solving this system of equations and taking into
account Eq. (1), we obtain
2.. 2 1/4
- 1 .L”.!.&... N
¢o--g /2( m?l%___mgl§ ) .. 14 rad/s.
1.76. We denote by ll and lz the lengths of the
sgrings connecting the axle to the first ball and
t e first and_the second ball. Since the balls move
in a circle, their equations of motion can be written in the form
mmzli = k (Z1 ··· Z0) -· k (Z1 -· $0),
mwz (Z1 + I2) = k (lz - %)»
whence
Z=
1 1-·3mo>2/k-}—(moJ2/k)2 ’
I __ (1—m¢o2/k)l0
2* 1—3mco2/k—|—(moJ2/k)2 '
The solution has a physical meaning when the following inequalities are satisfied:
Bmcoz , mcoz mm2
Let us suppose that mm2/lc = x. Since mon?/lc > 0,
the second condition implies that 0 < .1: < 1. The
first condition yields
:1:2 -— 3.1: —|—- 1 > O,
`whence either .1: > (3 —I— 1/5)/2 x 2.6, or 37 <
(3 — V -5)/ 2 Lz 0.4. Consequently, the region of
Solutions 177
admissible values of :4: lies between 0 and
(3--}/5)/2, whence
3 - 1/5 1,:
°’ < l/*7* 771.77. The change in the kinetic energy Wk of the
body as a result of a small displacement As can
be written in the form
AWR = F AS,
where F is the force acting on the body. Therefore,
the force at a certain point of the trajectory 1S de. Wk
I ‘ ‘ . II A
INEIIIIIIIWEI
II IIIIIIAM W ’
QIEIII
h:¤IIIIII
In ` 'EE}
6CAs
Fig. 178
fined as the slope of the tangent at the relevant
point of the curve describing the kinetic energy as
a function of displacement in a rectilinear motion.
Using the curve given in the condition of the problem, wep find that (Fig. 178) FC nv. -1 N and
F .2 -3 N.
128. The amount of liberated heat will be maximum if the block traverses the maximum distance
relative to the conveyer belt. For this purpose, it
is required that the velocity of the block relative
to the ground in the vicinity of the roller A must
2-0771
178 Aptitude Test Problems in Physics
be zero (see Fig. 41). The initial velocity of the
block relative to the ground is determined from
the conditions
—-v0—}—at=O, Z?-v0t’* ,
where a = pg is the acceleration imparted to the
block by friction. Hence
v¤= 1/2ugl·
The time of motion of the block along the
conveyer belt to the roller A is
g=_
P8
The distance covered by the block before it
stops is
s1=l+v¢=l—}—v .
Then. the block starts moving with a constant
acceleration to the right. The time interval in
which the slippagle ceases is 1 = v/a = v/pg. The
distance by whic the block is displaced relative
to the ground during this time is
S__ at? _ v2
— 2 M 2pg °
Since v < 1-/ 2pgl by hypothesis, the block does
not slip from the conveyer belt during this time,
i.e. s < l.
The distance covered by the block relative to
the conveyer belt during this time is
U2 I U2
;· --—·-— ’[' :.:-—·-—•
$2 2a U 2pg
Solutions 179
The total distance traversed by the block relative to the conveyer belt is
. ET vt (v+V?»i?i)’
s=s s=l v /—— ———=———·———-—-—
‘+” +l ug+2uz 2m; °
The amount of heat liberated at the expense of
the work done by friction is
m v 2 Z3
1.79. In the former case (the motion of the pipe
without slipping), the initial amount of potential
energy stored in the gravitational field will be
transformed into the kinetic energy of the pipe,
which will be equally distributed between the
energies of' rotary and translatory motion. In the
latter case (the motion with slipping), not all the
potential energy will be converted into the kinetic energy at the end of the path because of the
work done against friction. Since in this case the
energy will a so be equally distributed between the
energies of translatory and rotary motions, the
velocity of the lpipe at the end of the path will be
smaller in the atter case.
1.80. After the spring has been released, it is uniformly stretched. In the process, very fast vibrations of the spring emerge, which also attenuate
very soon. During this time, the load cannot be
noticeably displaced, i.e. if the middle of the
sgring has been displaced by a distance x in doing
t e work A, the entire spring is now stretched by
x. Therefore, the potential energy of the spring,
which is equal to the maximum kinetic energy in the
subsequent vibratory motion, is Wk=kx”/2, where
k is the rigidity of the entire spring. When the
spring is pulled downwards_at the midpoint, only
its upper half (whose rigidity is 2k) is stretched,
and the wo1·k equal to the potential energy of extension of the upper part of the spring is A =
2k (x2/2) = ka:2. Hence we may conc ude that the
maximum kinetic energy of the load in the subsequent motion is Wk === A/2.
12*
180 Aptitude Test Problems in Physics
1.81. Since the system is closed, the stars will
rotate about their common centre of mass in concentric circles. The equations of motion for the
stars have the form
Here wl and mz are the angular velocities of rotation of the stars, ll and Z2 are the radii of their
orbits, F is the force of interaction between the
stars, equal to Gmlmz/Z2, where l is the separation
between the stars, and G is the gravitational constant. By the definition of the centre of mass,
mlll = m2l2, ll —}- lz = Z. (2)
Solving Eqs. (1) and (2) together, we obtain
G(m —|—.m) _ G(m —l—m
o1=o2= ]/——-—-*ig———i—=l ‘ ]/—————--1, 2) ,
and the required period of revolution of these
stars is
T=2nl I/—-ii-,
G <mi+m—2>
1.82. Let U1 be the velocity of the station before
the collision, U2 the velocity of the station and the
meteorite immediately after the collision, m the
mass of the meteorite, and 10m. the mass of the
station.
Before the collision, the station moved around
a planet in a circular orbit of radius R. Therefore,
the velocity U1 of the station can be found from
the equation
10mv% _ G 10mM
R * R2 ‘
Hence vl = }/GM/R. In accordance with the momentum conservation law, the velocities u, ul,
and v2 are connected through the following relation:
mu + 10mvl = ffmv,.
Solutions 181
We shall write the momentum conservation law
in projections on the ax- and y-axes (Fig. 179}:
10mvl = 11mv,x, (1)
mu = 11mv,y. (2)
After the collision, the station goes over to an
elliptical orbit. The energy of the station with the
if
.2:
·<··· \
U3 X \x
ll `\
(1
II
:·P
I
\ M ll.
\`\ I] I
\ ll
ly
Fig. 179
meteorite stuck in it remains constant during the
motion in the elliptical orbit. Consequently,
11mM 11m
*6 -7;-+-2- (”§»+"Zy)
11mM 11m
-· -G-@7+7 W, (3)
where V is the velocity of the station at the momen t
of the closest proximity to the planet. Here we
have used the ormula for the potential energy of
gravitational interaction of two bodies (of mass
ml and m,): WD == —Gm1m,/r. According to
Kepler’s second law, the velocity V is connected to
182 Aptitude Test Problems in Physics
the velocity v of the station immediately after
the collision through the relation
VR
T=”2xR• (4)
Solving Eqs. (1)-(4) together and considering that
vi ··= 1/GM/R, we determine the velocity of the
meteorite before the collision:
58GM
u = `—R"`°
1.83. For a body of mass m resting on the equator
of a planet of radius R, which rotates at an angular velocity oa, the equation of motion has the
form
mco2R = mg' — N,
where N is the normal reaction of` the planet surface, and g' = 0.01gis the iree—fa1l aoce eration on
the planet. By hypothesis, the bodies on the equator are weightless, i.e. N = 0. Considering that
to = 2:n;/T, where T is the period of rotation of
the planet about its axis (equal to the solar day),
we obtain
T2
Rrzzr K'Substituting the values T = 8.6 X 10* s and
g' z 0.1 m s’, we get
R #.21.8 X 10" m = 18000 km.
1.84. We shall write the equation of motion for
Neptune and the Earth around the Sun (for the
sake of simplicity, we assume that the orbits are
circular):
GMm
mNm?qRN =··--T?§I—§— ,
GM mE
"·•7•*¤¤•·"T·* •
L RE I
Solutions 183
Here mN, mE, (0N, cog, RN, and RE are the
masses, angular velocities, and orbital radii of Nelptune and the Earth respectively, and M is t e
mass of the Sun. We now take into account the
relation between the angular velocity and the
period of revolution around the Sun:
.0., .. .2; ,, -22.
L1E•
Here TN and TE are the periods of revolution of
Neptune and the Earth. As a result, we fmd that
the period of revolution of Neptune around the
Sun is
YG?
TN z TE —·§····· 2 y€&l1°S.
RE
A similar result is obtained for elliptical orbits
from Kepler’s third __law.
'//'·"//>’ /_},}’/ / /7// ,'-/IY 5,
cz u ar 0
T, J *2 2
’”x9 NH mzg
Fig. 180
1.85. Let us consider two methods of solving this
problem.
1. The equilibrium conditions for the loads
have the form (Fig. 180)
T1 = ***18% Tg "· meg,
TICOSGI z T2C0Sd»a•
184 Aptitude Test Problems in Physics
in
From these relations, we can determine the angles
corresponding to the equilibrium position of the
system:
Sm <=¤1=—·—-—-—-’V’“Z·JJ‘,,€f’ "‘“‘ · m»=”——-——’“§$lf
Obviously, equilibrium can be attained only
under the conditions that O < oc, < crc/2 and 0 <
dg < It/2, i.€.
M?-—m§—|~m% M2-—m¥—}—m§
0<——-ETIE--—<1, 0< 2Mmz <1.
These inequalities imply that the entire system
will be in equilibrium only provided that
M<m1-l—m2, M2>|m§—m§|.
2. Let us consider the equilibrium of point A.
At this point, three forces are applied:
T1 = mls, T2 = mag, T3 = Ma
Point A is in equilibrium when T1, T2, and T, form
a triangle. Since the sum of two sides of a triangle
is larger than the third side, we obtain the relation
between the masses ml, mz, and M required for
the equilibrium of point A:
m1+m,>M,M—}-m1>m2, M—|—m2>m1.
1.86. Let us consider the equilibrium conditions
for the rod at the instant when it forms an angle on
with the horizontal. The forcesactingl on the rod
are shown in Fig. 181. While solving t is problem ,
it is convenient to make use of the equality to zero
of the sum of the torques about the point of intersection of the force o gravity mg and the force F
applied by the lperson perpendicular to the rod
(point 0) since t e moments of these forces about
this point are zero.
If the length of the rod is 21, the arm of the
normal reaction N is I cos cc, while the arm of the
Solutions 185
friction is I/sin on —{— lsin oc, and the equilibrium
condition will be written in the form
Nl cos ot: FUI (E3;-{—sin oc) =F;,l ,
whence
cos oc sin on cos ot sin ot
Ff"_N 1—|—sin°o¤ "_N 2sin°oz-|—cos2<z
1
_N 2tanot—l—cotoz °
On the other hand, the friction cannot exceed
the sliding friction pN, and hence
/ 2tanot—{—cotoz ’
This inequality must be fulfilled at all values of
the angle ot. Consequently, in order to {ind the
On
I `~\
{F
I
I
I
I
I
N /*9
0: I
Ffr/
Fig. 181
minimum coefficient of friction umm, we must find
the maximum of the function (2:;* -|- 1/:::2)*1, where
aca = tan oc. _The identity 2.1:* + 1/x2 =··- (1/2:c ———
1/:c)“ —I- Z}/Zimplies that the maximum value of
186 Aptitude Test Problems in Physics
1/(2 tan on + cot cz) is 1/(21/2) =-· ]/2/4 and is attained at ::2 = tan oc = 1/2/2. Thus, the required
minimum coefficient of friction is
V'?
pmin = "'Z" ·
1.87. Since the hinge C is in equilibrium, the sum
of the forces applie to it is zero. Writing the pro/_ __ U. » y,,{ _ _ ’./.f//I ’//
.4 _ ·
/ T T v,
L J // rghfllg F
-II
rm '”'mn}9\V\ \
/0\\
Fig. 182
jections of the forces (Fig. 182) acting on the hinge
C on the axis perpendicular to AC, we obtain
(m —l— mh",) g sin on = T cos ot, (1)
where mmn is the mass of the hinge. Similarly,
from the equilibrium condition for the hinie D
and fniom the lcondition that the middle rod is orizonta , we o tain
T cos on = F cos B —l— mmng sin oc. (2)
Solving Eqs. (1) and (2) together, we {ind that
T COS G"* sin G mg Sin G •
F·""""""`?;`Z»“`k3'§i"""""="`éY>§`(¥" >mgS"'°°‘
Solutions 187
Thus, the minimum force Fmm for which the 1niddle
rod retains its horizontal position is
Fmm=mgSlI1CZ = 1;and directed at right angles to the rod BD.
1.88. By hypothesis, the coefiicient of sliding friction between the pencil and the inclined plane
satisfies. the condition p. 9 tan ot. Indeed, the pencil put at right angles to the generatrix is in equilibrium, which means that mg sin on = FU, where
mg is the force of gravity, and Fr, is the force of
friction. But Ft, Q pmg cos oc. Consequently,
mg sin on { pmg cos ot, whence p > tan ot.
Thus, the pencil will not slide down the inclined
plane for any value of the angle qa.
The pencil may start rolling down at an angle
(po such that the vector of the force of gravity
¢2
to / ”
yy /
0A’
/
/
lla - /
Fig. 183
"leaves” the region of contact between the pencil and
the 1ncl1ned plane (hatched region in Fig. 183).
In order to_find this angle, we project the centre of
mass of the pencil (point A) on the inclined plane
and _mark the Hoint of intersection of the vertical
passing throug _the centre of mass and the inclined plane (point Obviously, points A and B
will be at rest for d1 erent orientations of the penC·1l1l1t8 centre of mass remains stationary. In this
188 Aptitude Test Problems in Physics
case, AB == 2l cos 30° tan or., where 2l is the side
of a hexagonal cross section of the pencil, and
2l cos 30° is the radius of the circle inscribed in
the hexagonal cross section.
As long as point B lies in the hatched region,
the pencil will not roll down the plane.
Let us write the condition for the beginning of
rolling down
—f-]?———=AB, or ——i—-—~=l l/'Eltanoz,
cos cpo cos cpt,
whence
1
= TCCOS .
(P0 a I/3tanoz
Thus, if the angle rp satisiies the condition
1 at
arc o ( -.7---) < < -— ,
C S `|/3 tance \(P \ 2
the pencil remains in equilibrium. The expression
for the angle cpo is meaningful provided that
tan cx > 1/ \/3. The fact that the pencil put parallel to the generatrix rolls down indicates that
tan on > 1/ 1/3 (prove this!).
9
¤>
.1:
Fig. 184
1.89. Let the cross section of the surface be described by the function y shown in Fig. 184.
Since the rod must be in equi 1 rium 111 any positnon,
Solutions 189
the equilibrium can be only neutral, i.e. the centre of mass of the rod must be on the same level
for any position of the rod. If the end of the rod
leaning against the surface has an abscissa (sc) the
ordinate (yo) of its other end touching the vertical
wall can be found from the condition
12-·=ly (w>—y0l°+=¤2. yu-- y lx) zh l/l°—=¤“·
Since the rod is homogeneous, its centre of mass is
at the midpoint. Assuming for definiteness that the
ordinate of the centre of mass is zero, we obtain
y¤+y (#3) :0
27
whence
l2.,_. 2
ll (xl: Zi: LTL .
Only the solution with the minus sign has the
physical meaning. Therefore, in general, the cross
section of the surface is described by the function
y (,1;): ,. .. JLE.;11 ,
2
where a is an arbitrary constant.
1.90. In the absence of the wall, the angle of deflection of the simple pendulum varies harmonically with a period T and an angular amplitude ot.
The projection of the point rotating in a circle of
radius oc at an angular velocity to = 2n/T performs the same motion. The perfectly elastic collision of the rigid rod with the wall at an angle of
deflection B corresponds to an instantaneous jump
from point B to point C (Fig. 185). The period is
reduced by At = 2y/co, where y -·= arccos (B/ot).
Consequently,
T1 = zu `_"EY" 1 .
cn co
19.1) Aptitude Test Problems in Physics
and the sought solution is
-1,}-:1--1- arccos-E-.
T at on
F
I at
7 ·"
Fig. 185
1.91*. Let the ball of mass m falling from a height
h elastically collide with a stationary horizontal
surface. Assuming that the time of collision of the
hall with the surface is small in comparison with
the time interval At between two consecutive collisions, we obtain
8
As a result of each collision, the mo_m_e_ntum of the
ball changes by Ap = 2mv = 2m_`|_{_%gh. Therefore,
the same momentum Ap = 2m]/2gh is transferred
to the horizontal surface in one collision.
In order to determine the mean force exerted by
the ball on the horizontal surface we consider the
time interval 1* > At. The momentum transferred to
the horizontal surface during the time ·c is
Visit
At 2 ]/2h./g
Consequentlg, the force exerted by the gumping
balls on the i orizontal surface and average during
Solutions 191
the time interval Ts can be obtained from the relation
. AP
Fm Z “‘·E";’”g·
By hypothesis, the mass M of the pan of the
balance is much larger than the mass m of the
ball. Therefore, slow vibratory motion of the balance pan will be superimposed- by nearly periodic
impacts of the ball. The mean force exerted by the
ball on the pan is Fm = mg. Consequently, the
required displacement Aa: of the equilibrium position of the balance is
mg
Ax I T •
1.92. The force acting on the bead at a certain
point A in the direction tangential to the wire is
F = mg cos ca, where on is the angle between the
tangent at point A and the ordinate axis (Fig. 186).
A i/
JUG
Fig. 186
For the length of the region of the wire from the
origin to the bead to vary harmonically, the force
F acting at point A must be proportional to the
length IA. But F Q mg, and IA increases indefinitely. Consequently, there must be a point B at
which the proportionality condition is violated.
This means that oscillations with the amplitude
IB cannot be harmonic. _
1.93. It follows from the equations of motion for
the blocks
mldl "—’ Fel, zmda = ——Fel,
192 Aptitude Test Problems in Physics
where Fel is the elastic force of the spring, that
their accelerations at each instant of time are
connected through the relation az = ——a1/2. Hence
the blocks vibrate in antiphase in the inertial reference frame iixed to the centre of mass of the
blocks, and the relative displacements of: the
blocks with respect to their equilibrium positions
are connected through the same relation as their
accelerations:
Axl
A$2'= ·····E··· .
Then
Consequentl , the period of small longitudinal oscillations oiy the system is
2m
T Zn nf 3k 1.94. Let us mark the horizontal diameter AB =
2r of the log at the moment it passes through the
equilibrium position.
Let us now consider the log at the instant when
the ropes on which it is suspended are deflected
from the vertical by a small angle on (Fig. 187).
In the absence of slippage of the ropes, we can
easily find from geometrical considerations that
the diameter AB always remains horizontal in the
process of oscillations. Indeed, if EF _I_ DK, FK =
2r tan on z 2rot. But BD rz FK/2 as ron. Consequently, LBOD on as was indicated above.
Since the diameter AB remains horizontal all
the time, the log performs translatory motion, i.e.
the velocities of all its points are the same at each
instant. Therefore, ·the motion of the log is synchronous to the oscillation of a simple pendulum
of length I. Therefore, the period of sm_all oscillations of the log is
-1*:-.2¤ l/-L,
8
Solutions 193
1.95. The period of oscillations of the pendulum
in the direction perpendicular to the rails is
fT
(l is the length of the weightless inextensible thread)
since the load Mis at rest in this case (Fig. 188).
.’_ .. ;-;`\ , I M
E —F . E
Z on •
I
JI
0I
,4 Z;.,Q:_ 6 i m
Fig._187 Fig. 188
The period of oscillations in the plane parallel
to the rails ("parallel" oscillations) can be found
from the condition that the centre of mass of the
system remains stationary. The position of the
centre Qof mass of the system is determined from the
equation mll == M (l — ll). Thus, the ball perfo1·ms oscillations with point Oremaining at rest
and is at a distance Z1 = Ml/(M —l— m) from point
0. Hence the period of "parallel" oscillations of
the pendulum is
t Ml
T :2 ’ -—-———— ,
2 H V (m +M> ez
Consequently,
.72.- ..___M
T1 _— m+1w •
1.96. The force exerted by the rods on theload is
F1 = 2Ftm cos oc, while the force exerted on the
13-0771
194 Aptitude Test Problems in Physics
spring is F, = 2Ft n sin cz (see- Fig. 48). According to Hooke’s law, F2 —·= (1.51 -— 21 sin oz) k,
where k is the rigidity of the spring. As a result ,
F1 = 1.5lk cot cx. — 2lk cos oz.
In order to determine the period of small oscillations, we must determine the force AF acting
on the load for a small change Ah in the height of
the load relative to the equilibrium position ho =
2l cos ono. We obtain
AF 1.5lk A (cot oe) —— 2lk A (cos cz),
where
Mm O,) ...(;£9.9£.°.&) M 2 .._.é§..,
dd q=CZ0 SH12 do
A (cos oc) = — sin oc,) Aon.
Consequently, since Ah = ——2l sin ono Au, we find
that
AF: --1.5k —{—2kl sin oc,) Acc
= —5kZ AOL== -5kAh
because sin 0:0 = 1/2.
The period of small oscillations of the load can
be found from the formula T ¢= 2n I/m/(5k), where
m is the mass of the load determined from the
equilibrium condition:
1.5kl cot occ —— 2kl cos 0:0 : mg,
1/5/2
m = —-—————
kl/g
Thus,
__ r 1/31
T- .... 2n V -T6-é-.
1.97. At each instant of time, the kinetic energy of
the hoop is the sum of the kinetic energy of the
centre of mass of the hoop and the kinetic energy
Solutions 195
O
of rotation of the hoop about its centre of mass.
Since the velocity of point A of the hoop is always
equal to zero, the two kinetic energy components
a1·e equal (the velocity of the centre of mass is
equal to the linear velocity of rotation about the
centre of mass). Therefore, the total kinetic energy
of the hoop is mvz (m is its mass, and v is the velocity of the centre of mass). Acco1·ding to the
energy conservation law, mv2 = mg (r — h A),
where h A is the height of the centre of mass of the
hoop above point A at each instant of time. Consequently, the velocity of the centre of mass of the
hoop is v = 1/g(r — h A). On the other hand, the
velocity of the pendulum B at the moment when
it is at a height h A above the- rotational axis A is
v == ]/ 2g (r —- h,A), i.e. is ]/2 times_larger. Thus,
the pendulum attains equilibrium 1-/2 times sooner
than the hoop, i.e. in
1
I Z`-: ·——::·· 2* S.
V2
1.98. It should be noted that small oscillations of
the load occur relative to the stationary axis AB
._ L"/,
,0,);*. .,
j ,/
L / ’ -.
Al a L 0
W_;,___.;......;..;.-. ...—-..,...-— U
’”9
Fig. 189
Fig. 189). Let DC _l_ AB. Then small oscillations
of the load are equivalent to the oscillations of a
simple pendulum of the same mass, but with the
13*
196 Aptitude Test Problems in Physics
length of the thread
Z’=LSlI1<1=·=L —-—-—-—-—-·—-· y ,
1/z¤+<L/2>=
and the free-fall acceleration
g’=gcosot=g—————————.w,
V !“+ (L/2)**
where L = AD.
Thus, the required period of small oscillations
of the system is
TW
T :2 I/*7*: -—·.
rt g 2n g
L99. In order to solve the problem, it is sufficient
to note that the motion of the swing is a rotation
nz "\
____ a
”"9
Fig. 190
about an axis passing through the points where
the ropes are fixed, i.e. the system is a "ti1ted simple pendu1um" (Fig. 190). The component of the
force of gravity mg along the rotational axis does
not influence the oscillations, while the normal
component mg sin on is in fact the restoring force.
Solutions 197
Therefore, using the formula for tl1e period of a
simple pendulum, we can write
/h
T #2** V
where
lilz lil:
h Z *17**--1* : f··——·*··—--— »
l ll +l§ !/¢2—|·b2
sin oz. == ——-B—-I/,,2+ b2 °
Consequently, the period of small oscillations of
the swing is
,.__. We
1 -.2:1; `/ ag .
1.100. The period ol a simple pendulum is inversely proportionalto the square root of the freeiull acceleration:
1
T OC —j··;· ,
Vg
Let the magnitude of the acceleration of the `lift
be a. Then the period of the pendulum for the
lift moving upwards with an acceleration a will be
up 1/z+<¤
and for the lift moving downwards with the same
acceleration
1
Tdown OC ·7·—<==
l 8""“
Obviously, the time measured by the pendulum
clock moving upwards with the acceleration a IS
proportional to the ratio of the time tm, of the
198 Aptitude Test Problems in Physics
upward uniformly accelerated motion to the period
Tup:
: ......
:,30 =-g-‘{§—;<¤f up 1/:+¤.
The time measured by the pendulum clock moving
downwards with the acceleration a is
I td ..._..
’down=F£`r0i Octdcwn l/6"“·
By hypothesis, the times of the uniformly accelerated downward and upward motions are equal:
tdmm = tu', = tl/2, where tl is the total time of
accelerated motion of the lift. Therefore, the
time measured by the pendulum clock during a
working day is
, ti / g+a ”g·—a)
:~2(], —————g +]/ g +:0.
Here to is the time of the uniform motion of the
lift. The stationary pendulum clock would indicate
: z :0 —|— :0.
It can easily be seen that the inequality ]/g —l— a -|Vg — a < 2}/g is fulfilled. Indeed,
( I/2+4+1/g—¤ )2=s-I- 1/s“·——¤“<1
2 VE 2g ‘
Hence it follows that on the average the pendulum
clock in the lift lags behind: t' < t, and hence the
operator works too much.
i.i0i. Pascal’s law implies that the pressure of a
gas in communicating vessels is the same at the
same altitude. Since the tubes of the manometer
communicate with the atmosphere, the pressure
in them varies with altitude according to the same
law as the atmospheric air pressure. This means
that the pressure exerted by the air on the liquid
in diHerent arms of the manometer is the same and
Solutions 199
egual to the atmospheric pressure at the altitude
o the manometer. Thus, the reading of the manometer corresponds to the zero level since there is
no pressure difference.
1.102. We choose the zero level of potential energy
at the bottom of the outer tube. Then the potential
energy of mercury at the initial instant of time is
l
W1 = 2SlPmer5' ( "i‘ ) Z PmergSl2·
The potential energy of mercury at the final instant
. gu p·*
hT
ii
r<——i‘$
Fig. 191
(the moment of separation of the inner tube,
Fig. 191) is (by hypothesis, Z > h)
h
WI = zsxpmerg (“%;‘)+S}*Pmer€ »
where .z is the level of mercury in the outer tube at
the moment of separation. This level can be found
from the condition of the constancy of the mercury
volume:
2S.23—{-Sh==·-ZSZ, ·‘13=Z—%.
The difference in the- potential energies is equal
to the sum ol the requires!. work A done by external
forces andthe work done by the force of atmospher-
200 Aptitude Test Problems in Physics
ic pressure acting on the surface of mercury in
the outer tube and on the upper (sealed) end of the
inner tube. The displacement of mercury in the
outer tube is Z -— x, the corresgonding work of the
force of atmospheric pressure eing p0S (l —- x) =
Pmcrz·$’h (G — ==)· _
The displacement of the sealed end of the 1nner tube is I- (l —|— x), the corresponding work
being — 0Sa: = —pmc,gSh:r.
Therefore, the required work of external
forces is
A=Wt"‘W1‘PmerHSh(l*2x)
3}
=Pmer8Sh(l ---,1}).
1.103. The pressure at the bottom ol the "_vertical"
cylinder isp = po + pwgh, where p,, is the atmospheric pressure,pw is the density of water, and g
is the free-fall acceleration. According-·to Pascal’s
»·• { g
`a
is
Fig. 192
law, the same pressure is exerted on the lower part
of the piston in the "horizontal" cylinder. The
total pressure of water on the part of the piston
separated from the lower part by a distance .1:
along the vertical is p — pgs: (Fig. 192). _
Let us consider the garts of t e piston in the
form of narrow (of wi th Aa:) horizontal strigs
S€p€ll'€\l·Q(l by Cqlla dlSl&II¢¢8 (I f1°0IIl US CQDH‘9• T 0
Solutions 201
force of pressure exerted by water on the upper
strip is
[P " Pwg (T 'l” a)l AS»
while the force of pressure on the lower strip is
lp - Pwr (r — ¢¤)] AS.
where AS is the area of a strip. The sum of these
forces is proportional to the area of the strip, the
proportionality factor 2 (p —— pgr) being independent of a. Hence it follows that the total force of
pressure of water on the piston is
(P "‘ Pwgrl mg = [Po + Pwg (h _ Tl] KrzThe piston is in equilibrium when this force is equal
to the force of atmospheric pressure acting on the
piston from the left and equal to pom?. Hence
h === r,
i.e. the piston is in equilibrium when the level of
water in the vertical cylinder is equal to the radius of the horizontal cylinder. An analysis of the
solution shows that this. equilibrium is stable.
1.104. The condition of complete submergence of
a body is
M ..2 pwvv
where M is the mass of the body, and V is its volume. In the case under consideration, we have
mcork‘1"m·aI lpcork ‘1‘ pal
Hence it follows that the minimum mass of the
wire is
___ Pal (Pw_Pc0rk) m N1 6m _
'”“‘ <pa—pw> pmt °°"" "‘ °°"‘
1.105. Obviously, in equilibrium, the sphere is at
a certain height Iz. above the bottom of the reser-
202 Aptitude Test Problems in Physics
voir, and some part of the chain lies at the bottom,
while the other part hangs vertically between the
bottom and the sphere (Fig. 193). By hypothesis,
we can state that the sphere is completely submerged in water (otherwise, nearly the whole chain
would hang, which is impossible in view of the
{ -... -.- —.` " "-Z`
*:i‘;f~:-12 ; ,f
i`.".-".-" I`¤'—
ir;. .;" ’—- ‘ ;i'i
Fig. 193.
large density of iron`). Then the height h can be
obtained from the equality of the total force of
gravity of the sphere and the hanging part of the
chain and the buoyant force acting on them:
h-—D/2 h-—D 2
(M+m————) g=pw(V +-2- ——l——) gl Piron I
Hence
D pwV—M
2 + m (1""Pw/Pironl m
The depth at which the sphere floats is H —— h =·-1.4 m.
1.106. The equilibrium condition for the lever is
the equality of the moments of forces (Fig. 194).
In air, these forces are the forces of gravity mlg
and mzg of the bodies. Since they are different,
their arms ll and I, are also different since
wish ==‘· mesh-
Solutions 203
When the lever is immersed in water, the force
of grav1ty_ is_ supplemyepfed by the buoyan of
water, which 1s_propor onal to the volume of the
lr I2
r} [br;
/77,9 mgg
Fig. 194
body. By hypothesis, F1 = F2, and therefore
(mlg -- F1) ll ak (mzg —-- F2) 1,.
Thus, the equilibrium of the lever will be violated.
1.107. The volume of the submerged part of each
box changes by the same amount AV = m/pw,
where m is the mass of the body, and pw is the density of water. Since the change in the level of water
in each vessel is determined only by AV and tl1e
vessels are identical, the levels 0 water in them
will change by the same amount.
1.108. Let the volume of the steel ball be V, and
let the volume of its part immersed in mercury be
V0 before water is poured and V1 after water covers
the ball completely. The value of V0 can be found
from the condition
Pstv = Pmer V0,
where pst and pm, are the densities of steel and
mercury. Since the pressure of water is transmitted
through mercury to the lower part of the ball, the
buoyant force exerted on it by water is
pw (V V,) g, where pw is the density of water,
while the buoyancy of mercury is pm€,.V1g. The
204 Aptitude Test Problems in Physics
condition of floating for the ball now becomes
p8tV = PHIBPVI + Pw —- V1)•
whence
V :._ .E*§.Z'.£.‘L. v_
1 Pmer"‘Pw
Thus, the ratio of the volumes of the parts of the
ball submerged in mercury in the former and latter
cases is
_‘f_Q_:_— pst pII`l€I` __: 1‘·—_p\V//PIDBT
V1 Pmer Pst‘-Pw 1* Pw/Pst °
Since p 8,-> pst, V0 > V1, i.e. the volume of the
part of ntlie ball immersed in mercury will become
smaller when water is poured.
1.109. The level of water in the vessel in which the
piece of ice floats is known to remain unchanged
{¤lI
*{ 1 I
`I
Fig. 195.
after melting of ice. In the case under consideration, we sha l assume that the level of water at the
initial moment (measured from the bottom of the
vessel) is ho, and that the level of the surface of
oil is h (Fig. 195). If the vessel contained only water, its level hl for the same position of the piece
of ice relative to the bottom of the vessel would
obey the condition
Solutions 205
Water formed as a result of melting ice has a
volume corresponding to the hatched region in the
figure. Since a part of this volume is above the
surface of water (hl > hg), the level of water rises
after melting of ice. On the other hand, since
hl < h, oil hlls the formed "hole", i.e. the total
level of liquid in the vessel falls.
1.110. Let x be the length of the part of the rod in
the tumbler, and y be the length.of its outer part.
Then the length of the rod is sr: —l— y, and the centre
of mass of ,_the rod is at a distance (x-|—y)/2 from
its ends and at a distance (y — :::)/2 from the outer
end. Tl1e condition of equilibrium is the equality
to zero of the sum of the moments. of force about
the brim of the tumbler.
, M 1 —x
<m.——n.> ¤:=—¢’i=—g—-—i.
where Fb = mbg = palgV is the force of gravity
of the ball, and F A = pwgV/2 is the buoyant force,
where V = (4/3);rtr°* is the volume of the ball.
The required ratio is
1.111. When the barometer falls freely, the force of
atmospheric pressure is no longer compensated by
the weight of the mercury column irrespective of
its height. As a result, mercury iills the barometer
tube completely, i.e. to the division 1050 mm.
1.112. In a vessel with a liquid moving horizontally with an accele1·ation a, the surface of the liquid
becomes an inclined plane. Its slope cp is determined from the condition that the sum of the force of
pressure F and the force of gravity mg acting on
an area element of the surface is equal to ma, and
the force of pressure is no1·mal to the surface. Hence
tan = il- ,
(P 8
206 Aptitude Test Problems in Physics
According to the law of communicating vessels, the
surfaces of the liquid in the arms of the manometer
belong to the above-mentioned inclined plane
(nga
·____, i `;1·.l.$0f#¤ lit; —
\
Fig. 196
(Fig. 196). It follows from geometrical considerations that
h —h
t :__ 2 I
‘“‘ "’ hz-thi
whence
G: 8 (h2""h1)
ha+hi `
1.113. The air layer of thickness Ax at a distance x
from the front of the cabin experiences the force of
pressure
[p[(¤= + Ap) -· p (¤¤>l 63
where S is the cross—sectional area of the cabin.
Since air is at rest relative to the cabin, the equation of motion for the mass of air under consideration has the form
pS Axa = [p (sc —|- Ax) —p (:1;)] S.
Making Ax tend to zero, we obtain
dp
”rE""’“·
whence
p (¤=> = pi + pax-
Solutions 207
Since the mean pressure in the cabin remains unchanged and equal to the atmospheric pressure po,
the constant pl can be found from the condition
l
P0: P1 'l` jgg"" 1
whgrlti Z Qs the lekngth gf the cabin. Thusi in tllie
mi e o t e ca in, t e pressure is equa to t e
atmospheric pressure, while in the front and rear
parts of the cabin, the pressure is lower and higher
than the atmospheric pressure by
Ap:-2gL 4; 0.02. Pa
respectively.
1.114. Let us consider the conditions of equilibrium for the mass of water contained fbetweeieln cross
sections separated by x and x + Ax rom t e rotational axis relative to the tube. This part of the
liquid, whose mass is pWS Ax, uniformly rotates at
an angular velocity no under the action of the forces
of pressure on its lateral surfaces. Denoting the
pressure in the section .1: by p (ac), we obtain
,A
[p(:c-{—A:1:)-—-p(x)]S==pS Axcoz (x—;——-ii:-) ,
Making Ax tend to zero, we obtain the following
equation:
::'pWm2xv
whence
$2
p(=¤)==pw<¤2 ‘l‘P0·
Using the conditions of the problem
2
P1=P("1l= Pwwz +P0»
2 ri
p2=p(r2) Z Pw<•> +P0»
208 Aptitude Test Problems in Physics
we obtain the angular velocity of the tube:
__ 2 P2"P1
"’ ‘ l/ Pw rs—»-2 ·
1.115. Let the_exponent on be such that the body
having an initial velocity v traverses a finite distance s (v) in qgthe medium. Since the velocity of
the body monotonically decreases during its motion
in the medium, s (v) > s(v1) for v > v,. It is also
clear that s (v) tends to zero as v-» 0. The condition under which the body comes to a halt is
that the work A of the drag is equal to the initial
kinetic energy of the body:
mvz
-2--- A. (1)
Since the drag monotonically decreases with the
velocity of the body during its motion, we can
write
A Q pvas (v). (2)
Substituting Eq. (2) into Eq. (1), we obtain
ll 2—¤¤
$(*02 2p v .
whence it follows that for cz 2 2, the condition
lim s (v) =· O is violated. Therefore, for on 2 2,
v-+0
the body cannot be decelerated on the final region
of the path.
1.116. According to the law of universal g1·avi—
tation, the force of attraction of the body of mass m
. . Mmm .
to Mars on ICS surface 18 G—§r, where MM is the
M
mass of Mars, and RM is its radius. This means
that the free-fall acceleration on the surface of
Mars is gM= GMM/RK;. If the mass of the Martian atmosphere is mM, it is attracted to the surface
of the planet with the force mMgM, which is equal
Solutions 209
to the force of pressure of the atmosphere, i.e. the
pressure on the surface of Mars is pM === mMgM/
(iixtlilii). Similarly, for the corresponding parameters on the Earth, we obtain PE = mEgE/(4¤tR§E).
The ratio ol the masses of the Martian and the
Earth’s atmospheres is
mM __ I’i»1·4¤R§;&’m
me T pE·4¤¤¤R§;sM
Considering that MM = (4/3)nRi'{IpM (a similar
expression can be obtained for the Earth) and substituting the given quantities, we get
;*s.:.m..fm.esE3_4X,O-3_
mn PE RE PM
It should be noted that we assumed in fact that
the atmosphere is near the surface of a planet.
This is rea ly so since the height of the atmosphere
is much smaller than the radius of a planet (e.g.
at an altitude of 10 km above the surface of the
Earth it is impossible to breathe, and the radius
of the Earth is RE rx 6400 km!).
2. Heat and Molecular Physics
2.1. Since the vertical cylinders are communicating
vessels, the equilibrium sets in after the increase
in the mass of the first piston only when it "sinks"
to the bottom of its cylinder, i.e. the whole of the
gas [lows to the second cylinder. Since the temperature and pressure of the gas remain unchanged,
the total volume occupied by the gas must remain
unchanged. Hence we conclude that Slho -l- Szho =
S h, where S1 and S are the cross-sectional areas
of? the first and secondz cylinders, and h is the height
at which the second piston will be located, i.e.
just the required difierence in heights (since the
first piston lies at the bottom). The initial pressures
produced by the pistons are equal. T ereiore,
mig :____ mzg
S1 S2 ’
fi ___ ml
S2 M mz
and hence
h=h0 l=0.3 m.
2.2. If the temperature Twau of the Vessel Walls
coincides with the gas temperature T, a molecule
striking the wall changes the normal comlponent
px of its momentum by -—px. Consequent y, the
total change in the momentum is 2px. When
Twau > T, the gas is heated. This means that gas
molecules bounce off the wall at a higher velocity
than that at which they impinge on the wall,
Solutions 211
and hence have a higher momentum. As a result,
tl1e change in the momentum will be larger than
2px (Fig. 197).
Ii, however, Twau < T, the gas is cooled, i.e.
gas molecules bounce off the wall with a smaller
momentum than that with which they impinge on
the wall. In this case, the change in the momentum
will be obviously smaller than 2px (Fig. 198).
{—¤•...;.¤.»-•»•·-• ·<·-I- l
Qi
”
Fig. 197 Fig. 198
Since according to N ewton’s second law, the change
in momentum is proportional to the mean force,
the pressure exerted by the gas on the walls is
higher when the walls are warmer than the gas.
2.3. The work A done by the gas during the cycle
is determined by the area of the p-V diagram
bounded by the cycle, i.e. by the area of the trapezoid (see Fig. 57):
V —-V V ———V ~
A=<p2.—-pi)(i———£—`;¥—·—#‘*-——¥All these quantities can easily be expressed in terms
of pressure and volume pl and V1 at point_1. Indeed,
according to Charles’s law, V3 = V2T3/T2 -·-=
VITS/T2 and V4 : VIT4/T1 == V17'2/T1, while the
{Gay-Lussac law implies that pz = p1T2/T1. Substituting these values into the expression for work,
we obtain
Z a 87 /.....T2"·T.1 ) ·_-?L.__»)
14*
212 Aptitude Test Problems in Physics
The equation of state for n moles of an ideal gas
is p1V1 = nR T1, and we can finally write
T.T
.4: R 1* -1* -1-2),
2.4. Figure 58 shows that on segments 1-2 and
3-4, pressure is directly proportional to temperature. It follows from the equation of state for an
ideal gas that the gas volume remains unchanged
in this case, and the gas does no work. Therefore,
we must find the work done only in isobaric processes 2-3 and 4-1. The work A2, = pz (V, — V,)
is done on segment 2-3 and A ,,1 = pl (V1 — V4)
on segment 4-1. The total work A done by the gas
during a cycle is
A = P2 (Vs ‘· V2) + p1(V1 _ V4)The equation of state for three moles of the
ideal gas can be written as pV = 3R T, and hence
mVi = 3RT1» PIV4 = 3RTn p2V2 = 3RT2.
p2V3 == p3V3 = 3RT¤·
Substituting these values into the expression for
work, we finally obtain
:3R(T1“l` T:a‘—T2"T4)
=2><10"·J=20kJ.
The cycle 1 -»- 4-> 3—» 2 -—» 1 is in fact equivE to two simple cycles 1 —> 0 -> 2 -+ 1 and
I —» 3 —+ 0 (see Fig. 59;; The work done by
as is determined by t e area of the corre-`ng cycle on the p-V diagram. In the first
‘.he work is positi·ve, while in the second
is negative (the work is done on the gas;.
·k done in the first cycle can easily be ca A _ (Po"P1) (Va“"V1)
1" 2 •
As regards the cycle 0 -> 4 —>- 3 —> 0, the triangle
on the p-V diagram corresponding to it is simi ar
Solutions 213
to the triangle corresponding to the first cycle.
Therefore, the work A, done in the second cycle
will be
(P2"Po)2
A =—A ———--———,
2 1 (P0""P1)2
(The areas of similar triangles are to each other as
the squares of the lengths of the corresponding elements, in our case, altitudes.) The total work A
done during the cycle 1 —> 4 —> 3 —>· 2 -> 1 will
therefore be
i—(p —p >“
A=A [—-L--9-] .»z 750 J.
1 (1¤»——pi>“
2.6. According to the first law of tnermodynamics,
the amount of heat AQ, received by a gas going
.0
pj _"""'“"""" 2
p. »... ,.
II
P0 ———— fl : :
{II
III
III
III
V0 IQ V2 V
Fig. 199
over from state 1 (p , V) to state 2 , V
(Fig. 199) is ° ° 1111 1)
A01 = AUI + Au
where AU, is the change in its internal energy,
and A, is the work done by the gas,
A __ (Po‘I·P1)(V1*V0)
1` "`___.‘E""""_` °
214 Aptitude Test Problems in Physics
As the gas goes over from state 1 to state 3 (p,, V,)
époints 2 and 3 lie on the same isotherm), the
ollowing relations are fulfilled:
AQ2=AU2‘l‘A2»
A :__: (170+ P2) (V2—Vol
22•
Since the final temperature of the gas in states 2
and 3 is the same, AU, = AU,. In order to find
out in which process the gas receives a larger
amount of heat, we must compare the works A1
and A,:
A __A __ (Po+P1) (V1"V0l ____ (Po+Pz) (V2“"Vol
1 “` """—`T""""" """"T"""`
_:_(p0V1" Povzl Q- (Pzvo ‘*· PIVD) < 0
since po V1 < p, V, and p,V, < p1V,. Consequently,
A, > A1 and AQ, > AQ1, i.e. the amount of heat
received by the gas in the process 1 —» 3 is larger.
2.7. Since hydrogen difiuses through all the partitions, it uniformly spreads over the entire vessel,
and in all the three parts of the vessel, the pressure
of hydrogen is
p _ mHg
(if _a gas penetrates through a partition, its pressure on both sides of the partition must be the
same in equilibriumgl.
Nitrogen can di use only through the right
partition, and hence will fill the middle and right
parts of the vessel (see Fig. 61) having the volume
(2/3)V. The pressure of nitrogen is
__ mw, BRT
PNB" ”N2 •
Solutions 215
Oxygen does not diffuse through the partitions,
and its pressure in the middle part of the vessel is
P _; m'O2
O2 NO2 V °
According to Dalton’s law, the pressure in a part
of a vessel is equal to the sum of the partial pressures of the gases it contains:
pl:-pH,) :2 1.3><10° Pa=1.3 GPa,
pz:-. pH2—}- pO2-|-pN2 g. 4.5 ><10° Pa: 4.5 GPa,
p3-=pH2—{—pN2 z 2.0 ><10° Pa-: 2.0 GPa.
2.8*. Let us first determine the velocity of the
descent module. We note that the change in pressure Ap is connected with the change in altitude
Ah through the following relation:
AP = —P€ Ah, (1)
where p is the gas density. The equation of state
for an ideal gas implies that p == (p/p.) RT (here
T is the gas temperature at the oint where the
change in pressure is considered) Taking into
account that Ah = —v At, where v is the velocity
of the descent, and At is the time of the descent,
we can write expression (1) in the form
Ap _ uv At
T_8 . (2)
Knowing the ratio Ap/At, i.e. the slope of the
tangent at the final point A oi the grap , we can
determine the velocity v from Eq. (2). (It should
be noted that since the left-hand side of (2) con»tains the ratio Ap/p, the scale on the ordinate
axis is immaterial.) Having determined (Ap! At) p‘1
from the graph and substituting p == 44 g/mol
216 Aptitude Test Problems in Physics
for CO2, we hud that the velocity of the descent
module of the spacecraft is
RT Ap
7:: '*"‘ '*""""
' an p Ar
8.3 J/(K·mol) ><7><102 K
— as 11.5 m/s.
10 m/s2><44>< 10 kg/mol >< 1150 s
Let us now solve the second part of the problem. Considering that the modulo has a velocity
of 11.5 m/s, it was at an altitude h = 15 km above
the surface of the planet 1300 s before landing,
i.e. this moment corresponds to t = 2350 s. Using
the relation (Ap/At) pr , we can find the required
temperature Th at this point of the graph from
Eq. (2):
: guv ( p_At ) N
Th ——-—·R ————Ap {..430 K.
lv (Ls
2.9. Since the piston has been displaced by h under
the action of the load, the volume of the gas has
decreased by hS and has become V —— hS. The gas
pressure under the piston is equal to the atmospheric
pressure po plus the pressure M g/ S produced y the
oad, i.e. po —}- Mg/S. Therefore, we can write
the equation of state for the gas before and after
loading:
p0V= HRT], (1)
MS
=nRr,. (2)
Here T, and Tr are the initial and final temperatures of the gas.
Since the gas is thermally insulated by hypothesis, it follows from the first law of thermody—
namics that the entire work A done on the gas is
spent to change its internal energy. i.e. A =
(3/2)nR (T; — T,) (the internal energy of a mole
of an ideal gas is U = (3/2)RT). It can easily
Solutions 217
be seen that the work is A = Mgh, and hence
3
Mgh=-5 HR (Tf—T]). (3)
Subtracting Eq. (1) from Eq. (2) termwise and
using expression (3) for T, —- T1, we obtain the
following equation in hz
WV
;‘%—·—Mgh— p0hS = é-Mgh. (4)
Hence we find that
h__ 17WgV
S (POS —l—Mt’/3) '
Substituting h into Eq. (2), we determine the final
temperature of the gas:
Tf: (POS +Mg> (31003 —2Mg> V
(3p0S—|—Mg) SnR '
2.10. According to the first law of thermodynamics,
the amount of heat Q supplied to the gas is spent
on the change AU in its internal energy and on
the work A done by the gas:
Q = AU —\~ A.
The internal energy U of a mole of an ideal gas can
be written in the form U -·= cVT = (3/‘2)R T, i.e.
AU = (3/2)R AT. The work done by the gas at
constant pressure p is A = p AV = pS Ax, where
Ax is the displacement of the piston. The gas pressure is
M
p 1- P0 + Tg 1
i.e. is the sum of the atmospheric pressure and the
pressure produced by the piston. Finally, the equation of state pV ·-= RT leads to the relation between the change AV in volume and the change A\T
in temperature at a constant pressure:
p AV = R AT.
218 Aptitude Test Problems in Physics
Substituting the expressions for AU and A
into the first law of thermodynamics and taking
into account the relation between AV and AT,
we obtain
35
Q=pAV—|—-5pAV=-§—pSAx. (1)
Since the amount of heat liberated by the heater
per unit time is q, Q = q At, where At is the corresponding time interval. The velocity of the
piston is v = Ax/At. Using Eq. (1), we obtain
,,:2. ....2....
5 Pos `l‘ M8 °
2.11*. For a very strong compression of the gas,
the repulsion among gas mo eeules becomes significant, and iiniteness of theirsize should be taken
into account. This means that other conditions
being equal, the pressure of a real gas exceeds
the pressure of an ideal gas the stronger, the larger
the extent to which the gas is compressed. Therefore, while at a constant temperature the product
pV is constant for an ideal gas, it increases with
decreasing volume for a rea gas.
2.12*. Let us consider an intermediate position
of the piston which has been displaced by a distance y from its initial position. uppose that the
gas pressure is tp, in the right part of the vessel
and pl in the le t part. Since the piston is in equilibrium, the sum of the forces acting on it is zero:
(P2 "" P1) S "" 2kl/ = O» (1)
where S is the area of the piston.
The total work done by the gas over the next
smal1_ displacement Ay of the piston is AA =
AA, + AA2, where AA2 is the work done by the
gas contained in the right part, and AAI is the
work done by the gas in the left part, and
Mi + Mi = pi AyS —- pi AyS
= (pi Z- pi) MS = 2ky Ay- (2)
Solutions 219
Thus, by the moment of displacement of the
piston. by x = Z/2, the total work done by the gas
will be equal to the sum of the potential energies
stored in the springs:
kZ2
A-2 a (T) <3>
If an amount of heat Q is supplied to the gas
in the right part of the vessel, and the gas in the
left part transfers an amount of heat Q' to tl1e
thermostat, the total amount of heat supplied
to the system is Q - Q', and we can write (the
first law of thermodynamics)
,_ k l 2
Q‘·Q —2 ·l·AU» (4)
where AU is the change in the internal energy of
the gas. Since the piston does not conduct heat,
the temperature of the gas in the left part does not
change, and the change AU in the internal energy
of the gas is due to the heating of the gas in the
right part by AT. For n moles of the ideal gas, we
have AU = n (3/2)R AT. The temperature increment AT can be found from the condition of equilibrium at the end of the process.
In accordance with the equation of state, the
pressure of the gas in the right part of the vessel is
p = nR (T + AT)/[S (l —l- l 2)]. On the other hand,
it must be equal to the sum of the gas pressure
p' = nRT/[S (l — l/2)] in the left part and the
pressure p" = 2kl/(2S) created by the springs, i.e.
2nR(T+AT) _ 2nRT +J_¢_i_
3Sl — Sl S
Hence we can find that AT = 2T —l— 3kl2/(2nR).
Using Eq. (4), we finally obtain
Q, :=Q——3Il-RT --2- fflz.
2.13. Let T1 be the initial temperature of the gas
under the piston, and T2 the gas temperature after
the amount of heat AQ has been supplied to the
220 Aptitude Test Problems in Physics
system. Since there is no friction and the vessel
is thermally insulated, the entire amount of heat
AQ is spent on the change AW in the internal energy of the system:
AQ = AW.
The change in the internal energy of the system
is the sum of the changes in the internal energy
of the gas and in the potential energy of the compressed spring (since we neglect the heat capacity
of the vesse , piston, and spring).
The internal energy of a mole of an ideal monatomic gas increases as a result of heating from
T1 to T, by
3
AW1=E-R(T2—T1). (1)
The potential energy of the compressed spring
changes by
k
AWz=+_.;(¤§—¤¤%>, (2)
where k is the rigidity of the spring, and xl and zz
are the values of the absolute displacement (deformation) of the left end of the spring at temperatures T1 and T respectively. Let us find the
relation between tlie parameters of the gas under
the piston and the de ormation of the spring.
h The equilibrium condition for the piston implies
t at
_ F _ kx _ pS
p""’ S “"3»"` 1 2:* k 1
where p is the gas pressure, and S is the area of the
piston. According to the equation of state for an
ideal gas, for one` mole we have pV = RT. For
the deformation x of the spring, the volume of the
gas under the piston is V = xS and the pressure
p = RT/(:cS). Substituting this expression for p
into Eq. (3), we obtain
s RT
$2 = T . (4)
Solutions 221
Thus, the change in the potential energy of the
compressed spring as a result of heating of the
system is
R
AWz= ·2·<Tz——T1)·
The total change in the internal energy of the
system as a result of heating from T2 to T2 is
AW = AW, —}— AW2 = 2R (T2 .. T1),
and the heat capacity of the system is
AQ __ AU _
C ·· rrr ztiqrr 2R·
2.14. Let us analyze the operation of the heat
engine based on the cycle formed by two isotherms
and two isochors (Fig. 200). Suppose that the temp s{2
`\ 2
G O2=A2
`s
..0/
4
V
Fig. 200
perature of the cooler (corresponding to the lower
isotherm)· is T1, and the temperature of the heater
(corresponding to the upper isotherm) is T2. On
the isochoric segment 1-2, the gas volume does not
change, i.e. no work is done, but the temperature
increases from T2 to T2. It means that a certain
amount of heat Q1 is supplied to the gas. On the
isothermal segment 2-3, the internal energy of
222 Aptitude Test Problems in Physics
the gas remains constant, and the entire amount
of heat Q2 supplied to the gas is spent on doing
work: Q2 = A .
On the isochoric segment 3-4, the gas temperature returns to its initial value T4, i.e. the amount
of heat Q4 is removed from the gas. On the isothermal segment 4-1, the work done by the gas
is negative, which means that some amount of
heat is taken away from the gas. Thus, the total
amount of heat supplied to tie gas per cycle is
Q1 —l— A2. Figure 200 shows that the work done by
the gas pe1· cycle is the sum of the positive work
A2 on the segment 2-3 and the negative work A4
on the segment 4-1.
Let us compare the pressures at the points
corresponding to equal volumes on the segments
4-1 and 2-3. The Gay-Lussac law indicates that
the ratio of these pressures is T4! T2, and hence the
work done by the gas is A4 ==- —-—(T4/T2) A2. The
total work per cycle is given by
A=Az+A4=(i- ?—1)A..
T2
and the efficiency is
__ A __ 1—T1/T2 4 T4
ln Q1+Az 1+Qi/Az < Ta °
Therefore, the efficiency of the heat engine based
on the cycle consisting of two isotberms and two
isochors is lower than the efficiency 1 —— T4/T2
of Carnot’s heat engine.
2.15*. Let us first determine the free-fall acceleration gpl on the surface of the planet. On the one
hand, we know that the force of attraction of a
body of mass m to the planet is mgm. On the other
hand, it follows from the law of universal gravitation that this force is GmM/rz, where G is the
gravitational constant. Hence we obtain gpl =
GM/rz. The pressure p exerted by the atmospheric
column of height h on the surface of the planet is
P = pgpih. (1)
Solutions 223
where p is the density of the atmosphere. While
determining the pressure of the atmospheric column, we assume that the free—fall acceleration is
independent of altitude. This assumption isjusti—
lied since by hypothesis the height of the atmosphere
is much smaller than the radius r of the planet
(I1. < r).
Using the equation of state for an ideal gas of
mass M occupying a volume V in the form pV =
(M/lt) HT and considering that p == M/V we
find that
..I1Z ·—£E'··
RT °
Substituting this expression for p into Eq. (1)
and cancelling out p, we determine the temperature T of the atmosphere on the surface of the
planet:
“ R n Hr? °
2.16. We must take into account here that the heat
transferred per unit time is proportional to the
temperature difference. Let us introduce the following notation: Touu, Touta and Tu, Tr, are the
temperatures outdoors and in the room in the hrst
and second cases respectively. The thermal power
dissipatedby the radiator in the room is kl (T- TI.),
where kl is a certain coefficient. The thermal
power dissipated from the room is Ic, (T,. -- Tout),
where kg. is another coefficient. In thermal equilibrium, the power dissipated `by the radiator is
equalto thepower dissipated from the room. Therefore, we can write
-_— Tl`1) Z k2 (TI`1 —— TO.l1t1)°
Similarly, in the second case,
ki (T —·· Tm) = #2 (Tm - Tam)Dividing the first equation by the second, we obtain
TT"‘TI`1 Z Tr1""Tout1
T··Trz Trz·Tootz °
224 Aptitude Test Problems in Physics
Hence we can determine T:
T T 4; —-T T t
T; I`2 OU I I']. OU 2 :60OC.
Trz+ Tout1···Toutz*·· Tri
2.17. The total amount of heat q liberated by the
space object per unit time is proportional to its
volume: q = otR3, where ot is a coefficient. Since
the amount of heat given away per unit surface
area is proportional to T4, and in equilibrium, the
entire amount of the liberated heat is dissipated
into space, we can write q = BR2T’* (the area of
the surface is proportional to R2, and fi is a coefficient). Equating these two expressions for q,
we obtain
oz.
T4: -—- R.
B
Consequently, the fourth power of the temperature
of the object is proportional to its radius, and
hence a decrease in radius by half leads to a_de—
crease in temperature only by a factor of V 2 cx
1.19.
2.18*. For definiteness, we shall assume that the liquid flowing in the inner tube 2 is cooled, i.e. T 12 >
TM, and hence Tu < TH. Since the cross-sectional area 2S -— S = S of the liquid _flow in
the outer tube 1 is equal to the cross-sectional
area S of the liquid flow in the inner tube 2, and
their velocities coincide, the decrease in temperature of the liquid flowing in tube 2 from the entrance
to the exit is equal to the decrease in temperature
of the liquid flowing in tube 1. In other words,
the temperature difference in the liquids remains
constant along the heat exchanger, and hence
Ti2_Tf1=Tf2_Ti1' (1)
In view of the constancy of the temperature
difference, the rate of heat transfer is constant
along the heat exchanger. The amount of heat Q
transferred from the liquid flowing in tube 2 to
the liquid flowing in tube 1 dur-ing ra time t is
Q = S1atkt(T1¤ ·· Tn)- _ (2)
Solutions 225
Here Slat is the lateral surface of the inner tube,
Slat = 2;nrl, where r is the radius of the inner
tube, i.e. str2 = S, and T == The amount
of heat Q is spent for heating the liquid flowing in
tube 1. During the time t, the mass of the liquid
flowing in the outer tube 1 is m = pvtS, and its
temperature increases from Tu to Tn. Consequently,
Q = 1>v¢$¢ (Tn — Tn)- (3)
Equating expressions (2) and (3), we obtain
TS"
2n I/71* U1(T12·—Tn)= PUSC (Tt1·" T11)Hence we can find TH, and using Eq; (1), Tm as
well. Therefore, we can write
, 1 2 "7`szk -1
Tr1=T12+(T11·—T1g)
1 2 Hflsizk -1
Tf2=Tu+(Ti2——Tu)
pvc
2.19*. As a result of redistribution, the gas plressure
obviously has the maximum value at t e rear
(relative to the direction of motion) wall of the
vessel since the acceleration a is imparted to the
gas just by the force of pressure exerted by this
wall. We denote this pressure by pmax. On the other
hand, pmax < psat. Considering that psa, > p
and hence neglecting the force of pressure exerted
by the front wall, on the basis of Newtonls second
law, we can write
mgasa
Pmax = T < Psat»
where mgas is the mass of the substance in the
gaseous state contained in the vessel. Consequently,
fora <pSatS/M, no condensation will take place,
while for a > pSatS/M, the mass of the gas will
15-0771
226 Aptitude Test Problems in Physics
become m = pSatS/a, and the vessel will contain
a liquid having the mass
]nHq::.:]l/I-fn-;-_ •
2.20. Boiling of water is the process of intense
formation of steam bubbles. The bubbles contain
saturated water vapour and can be formed when
the pressure of saturated water vapour becomes
equa to the atmospheric pressure (760 mmHg, or
105 Pa). It is known that this condition is ful lled
at a temperature equal to the boiling point of
water: Tboil = 100 °C (or 373 K). By hypothesis,
the pressure of saturated water vapour on the
planet is po = 760 mmHg, and hence the temperature on the planet is T = Thou = 373 K. Using
the equation of state for an ideal gas
pz Pol*
RTb¤11 ’
where p. is the molar mass of water, and po is the
atmospheric pressure, and substitutin the numerical values, we obtain p = 0.58 kg?m“.
2.21. When we exhale air in cold weather, it is
abruptly cooled. It is well known that the saturated
vapour pressure drops upon cooling. Water vapour
contained in the exhaled air becomes saturated
as a result of cooling and condenses into tiny water
drops ("fog”).
If we open the door of a warm hut on a chilly
day, cold air penetrating into the hut cools water
vapour contained in the air of the hut. It also
becomes saturated, and we see "fog", viz. the drops
of condensed water. `
2.22*. It is easier to solve the problem graphically.
The total pressure p in the vessel is the sum of the
saturated water vapour pressure phat and the
pressure of hydrogen pH:. According to the equa-
Solutions 227
tion of state for an ideal gas, the pressure of hydrogen is
___ mg, RT__ 2><10"’ kg><S.3 J/(mo1·K) T
puzm pH V I-- 2><·l0"‘” kg/mol><2>;1()‘3 1113
I
=-·.4.15><103T,
where PH;2 is measured in pascals. The pH: (T)
dependence is linear. Therefore, having calculated
pH2 (T) for two values of temperature, say, for
T1: 373 K, pH, =15.5><105 Pa,
T2=453 K, pHz=18.8><1O5 Pa,
we plot the graph of PH, (T).
Using the hint in the conditions of the problem,
we plot the graph of the function psat (T). "Com—
p, 705 Pa
~_______——__-____
a
/8.8 ..__....._. --- ' #0+12
/§3*::;;i;:—· :.
" } { ¤ /p$¤¢
I1
lI·
H/r{
IIII
275 4*4 575` l /2525 475. 7}/<
Ts .1*0 Q=440
Fig. 201
posing" the graphs of PH2 (T) and psat (T), we
obtain the graph of the temperature dependence
of the total pressure in the vessel, p (T) (Fig. 201).
Using the p (T) curve, from the initial and iinal
values of pressure specified in the conditions of the
15*
228 J Aptitude Test Problems in Physics
problem, we obtain the initial and final temperatures in the vessel:
p, =-— 17 >< 105 Pa, T, = T,z 380 K,
p, = 26 >< 105 Pa, T2 = T, z 440 K.
Let us now determine the mass of evaporated
water. Assuming that water vapour is an ideal gas,
we calculate the initial pv, and final p,.2 pressures
of water vapour in the vessel. For this purpose,
we make use of the obtained graphs. For T, ==~
380 K, the pressure of hydrogen is ph, cz 15.5 ><
105 Pa, and
pv1=p,-p,,,, as 1.5>< 105 Pa.
For T2 == 440 K, p},2 as 18 >< 105 Pa, and
[Jv2=pf—piIz L!. SXIOS Pa.
Let us write the equations of state for water
vapour at p,,,, T, and p,,2, T2:
pm = i'-*2- RTL pty: —”$’—5- RTW
llv PV
where mv, and m,,2 are the initial and linal masses
of vapour in the vessel. Hence we can determine
the mass of evaporated water:
V
Am :”"’5°"""“ Z hh"` (‘prY2‘2‘ " pry; l
__ 18 >< 10*3 kg/mol >< 2 >< 10·3m3
— 8.3 J/(K- mol)
8 Pa 1.5 Pa . _
2.23. Ii‘h is the height of water column in the
capillary, the temperature of the capillary, and
hence of water at this height, is
"~=·Z·`i“i’·}5·
Solutions 229
Water is kept in tl1e capillary by surface tension.
If oh is the surface tension at the temperature Th,
we can write
E2
pwgr
wl1ere pw is the density of water. Hence we obtain
U =··—"g'” =(-—‘”"" ) (£L)·
h 2 2 Tup °
Using the hint in the conditions ol the problem,
we plot the graph of the function o (T). The tem—
perature Th on the level ol the maximum ascent
. G,/uN/m
80 —
60 6(})
40 — I
.. Tx\l
-L/Yl r/ K l
20 - (log' ,
n
-.- e__1..-.
0 20 40 60 80
Fig. 202
of water is determined by the point of intersection
of the curves describing the (pgrl/2) T/Tm, and
o (T) depemlences. Figure 202 shows that Th as
80 °C. Consequently,
hz EB- cx 6.4 cm.
Tup
The problem can also be solved analytically
we note that the o (T) dependence is practically
inear.
230 Aptitude Test Problems in Physics
2.24, The condition of equilibrium for the soap
bubble film consists in that the air pressure pbubl
is the sum of the external pressure pl and the excess
pressure 40/ry due to surface tension. It should be
noted that there are two air—soap film interfaces
in the soap bubble, each of which produces a pressure 20/r. For this reason, the excess pressure is
2 X 20/r = 40/r. Therefore, we can first write the
equilibrium condition in the form
40
pbum=pi+;—.
After the radius of the bubble has been reduced
by half, the pressure produced by surface tension
becomes 80/r. By hypothesis, the temperature is
maintained constant, and hence (according to
Boyle’s law) a decrease in the volume of the bubble
by a factor of eight (its radius has decreased by
half) leads to an eight-fold increase in the air pressure in the bubble (it becomes Spbum), so that we
can write
80
8pbUb].= p2-I- T •
Substituting pbub, into this formula from the first
equation, we can finally write
240
P2: 81*1+ ···;_··· .
2.25. In the fireplace, large temperature gradients
may take place. If the bricks and the mortar are
made of different materials, ix; materials with
different temperature expansion coefhcients, the
fireplace can crack.
2.26. Let us su pose that the temperature of the
mixture of the fiquids having the initial temperatures T1 and T2 has become T. Since the vessel
containing the mixture is thermally insulated
(AQ = 0), we can write
c1m'1(T ‘· T1) + ¢zmz (T _ Te) = 0,
Solutions 231
whence
ELM. E2. --- _ -1
,,,2 ,1 )<r Ta on r> .
By hypothesis, 2 (Tl — T) = T1 —— T2, and
hence T -— T2 = T1 -— T, and the ratio (T -—T2)/
(T1 — T) = 1. Therefore,
Ln;];$Z£¤L
mz ’
i.e. the ratio of the masses of these liquids is inverse to tl1e ratio of their specific heats.
2.27. In the former case, the water in the test tube
is mainly heated due to convection since warm
water is lighter than cold water. In the latter case,
water is cooled only as a result of heat exchange
between water layers in the test tube. Since the
conditions of heat exchange between the test tube
and outer water remain the same, tl < tz.
It should be noted that if we change the parameters of the problem (20 °C -» 0 °C and 80 OC —>
4 °C), we shall obtain a reverse answer. The reason
lies in the anomaly of water. In the temperature
interval from 0 to 4 °C, cold water is lighter than
warm water.
2.28. For the system under consideration (vesselwater, vessel-water-ball), the heat flux per unit
time q = AQ/At through the surface of contact
with the ambient depends on the temperature
difference:
A
"AQ£":O*FlTves—T)»
where t is the time, TVCS is the temperature of the
vessel, T is the temperature of the ambient, and
F is a certain function of temperature. The coefficient oz. is determined by the conditions at the
contact of the system under consideration with
the ambient. In our case, the conditions at the
contact are identical for the two vessels, and hence
the coefficient cx is thesame in both cases. The
232 Aptitude Test Problems in Physics
amount of heat AQ lost by the vessel leads to a
decrease in the temperature of the vessel by ATVES.
QC For the*vessel with water, we obtain
AQ1 = (Mwcw 'l` mvescves) ATves»
where MW and cw are the mass and specific heat of
water, mvcs and cves are the relevant quantities
for the vessel.
For the vessel with water and the ball, we obtain
AQ2 = (Mwcw `l` mwcw `l` mbcbl ATves»
where mb and cb are the mass and specific heat of
the ball. By hypothesis, mvgs < MW and mb -=
MW. Besides, cves < cw, and hence we can write
AQ1 = Mwcw ATves• AQ2 = ·Mw (cw `l” Cb) ATves·
It can easily be seen that the change ATVBS of temperature in the two vessels occurs during different
time intervals At, and At, so that
ATves = (I Atl
F(Tves·‘T) Mxvcw ,
ATves ____ Q
F(Tves—T) Mw (¢w+¢b) 2`
Hence we obtain
At] __ CW
M2 — €w‘l‘Fb •
Therefore, the following relation will be fulfilled
for the total times tl and t2 of cooling of the vessels:
j_-L: vw —l-cs: k
tl CW ,
whence cb/cw = k — 1.
2.29. If the level of water in a calorimeter has
become higher, it means that a part of water has
been_frozen (the volume of water increases during
freezmg). On the other hand, we can state that some
amount of water has not been frozen since otherwise
nts volume would have increased by a factor of
Solutions 233
pw/picc z 1.1, and the level of water in the calorimeter would have increased by (h/3) (1.1 —- 1) z
2.5 cm, while by hypothesis Ah = 0.5 cm. Thus,
the temperature established in the calorimeter
is 0 °C.
Using this condition, we can write
= _—7" Am + clC€mlC€ OC -—-. TlC€)7
where Am is the mass of frozen water, and Tice
is the initial temperature of ice. As was mentioned above, the volume of water increases as
a result of freezing by a factor of pw/pice, and hence
Ah5=(.9£...4)&’ (2)
Pice Pw
where S is the cross—sectional area of the cilorin;5
eter. Substituting Am from Eq. (2) into q. (
and using the relations mw = (h/3) pwS and mice =
(h/3) p,ceS, we obtain
h
CWS '§` PWTW
, h.
Hence
T _ __ 7t 3Ah pw
icc cice h pw "‘ pics
_2&'.. TW §_54¤(;_
Cice Pice
2.30*. (1) Assuming that water and ice are incompressible, we·can find the decrease in the temperature of the mixture as a result of the increase in the
external pressure:
AT=(%-L) X 1Kz0.18K.
Such a small change in temperature indicates that
only a small mass of ice will melt, i.e. Am < mice,
234 Aptitude Test Problems in Physics
We write the energy conservation law:
4 = 71. Am — (cme -|- cw) m AT.
Let us estimate the work A done by the external
force. The change in the volume of the mixture
as a result of melting ice of mass Am is
AV =.£”.L.....€\L".
Pice Pw
: Am -P-Y’-3I‘££9. (0.1-TE-— .210-4 1113.
Pwplce Pice
We have taken into account the fact that the density of water decreases as a result of freezing by
about 10%, i.e. (pw —- pim)/pw as 0.1. There ore,
we obtain an estimate
A { pl AV = 0.25 kJ.
The amount of heat AQ required for heating the
mass m of ice and the same mass of water by AT is
AQ = (cicc —|— cw) m AT 911.1 kl.
Since A < AQ,. we can assume that 71, Am ·-= AQ,
whence
Am Z 'ég" =-' 3.2 gA.
The change in volume as a result of melting ice
of this mass is
AVI = Am —QiY-—`;&9£~ 24 3.5 X 10·" 1118.
Pwpice
Considering that for a slow increase in pressure the
change in the volume AV, is proportional to that
in the pressure Ap, we can lind the work done by
the external force:
A Z Bxgll A 0.44 J.
Solutions 235
(2) ·We now take into account the compressibility of water and ice. The change in the volume
of water and ice will be
AV' Z-: gé-10-2VOw + -if-;%10·2V0icc ex 2 >< 10‘° m3,
where V0W=1O‘° mi-’• and VOICE = 1.1 X 10*3 m3
are the initial volumes of water and ice.
The work A' done by the external force to compress the mixture is
I
.4* = ifi- is 2.5 J.
2
The total work of the external force is
Atot-=A +A' x3 J.
Obviously, since we again have Amt < AQ, the
mass of the ice that has melted will be the same
as in case (1).
2.31. Considering that the gas density is p = M/ V,
we can write the equation of state for water vapour
in the form p = (p/pt) RT, where p and p. are the
density and molar mass of water vapour. Boiling
takes place when the saturated vapour pressure
becomes equal to the atmospheric pressure. If
the boiling point of the salted water has been raised
at a constant atmospheric pressure, it means that
the density of saturated water vapour must have
decreased. _
2.32. Let us consider a cycle embracing the triple
point: A ->- B -» C -> A (Fig. 203). The following
phase transitions occur in tum: melting -» vaporization -» condensation of gas directly into the
solid. Provided that the cyc e infinitely converges
to the triple point, we obtain from the first law
of thermodynamics for the mass m of a substance
ml ·-}— mq — mv = 0
since the work of the system during a cycle is zero,
there is no inflow of heat from outside, and hence
the total change in the internal energy is also
equal to zero (t e right-hand side of the equation).
236 Aptitude Test Problems in Physics
Hence we can {ind the latent heat of sublimation
of water at the triple point:
V = Pt —l- q = 2.82 X 106 J/kg.
P
Liquid
Solid
A
ph- `"‘ `'*_ 6
0 { Gas
I
Fig._203
2.33. The concentration of the solution of sugar
poured above a ho1·iz0ntal surface practically re#
mains unchanged.
After the equilibrium sets in, the concentration
of the solution in the vessel will be
.ch
c = ———2 1 ,
The concentration changes as a result of evaporation
of water molecules from the surface (concentration
increases) or as a result of condensation of vapour
molecules into the vessel (concentration decreases).
The saturated vapour pressure above the solution in
the cylindrical vessel is lower than that above the
solution at the bottom by Ap == O.05pSat (c--c ).
This difference in pressure is balanced by the
pressure of the vapour column of height hz
pvgh =—= O.O5pSat (c -—- c2).
Hence we obtain
h:___ 0·05Psat @*62)
eva °
Solutions 237
The density pv of vapour at a temperature T =
293 K can be found from the equation of state for
an ideal gas:
Psatll
Pv RT .
Thus, the height h of vapour column satisfies the
quadratic equation
0.05c2RT 2h,
hz ———————— ———-—
Mg h—1
Substituting the numerical values and solving the
quadratic equation, we obtain
h 2 16.4 cm.
It is interesting to note that, as follows from
the problem considered above, if two identical
vessels containing solutions of different concentrations are placed under the bell, the liquid will
evaporate from the solution of a lower concentration. Conversely, water vapour will condense to
the solution of a higher concentration. Thus, the
concentrations tend to level out. This phenomenon
explains the wetting of sugar and salt in an atmosphere with a high moisture content.
2.34. Since the lower end of the duct is maintained
at a temperature T1 which is higher than the melting
point of cast iron, the cast iron at the bottom will
be molten. The temperature at the interface between the molten and solid cast iron is naturally
equal to the melting point Tm lt.
Since the temperatures of the upper and lower
ends of the duct are maintained constant, the
amount of heat transferred per unit time through
the duct cross section must be the same in any
region. In other words, the heat flux through the
molten and solid cast iron must be the same (the
brick duct is a poor heat conductor so that heat
transfer through its walls can be neglected).
The heat flux is proportional to the thermal
conductivity, the cross-sectional area, and the
temperature difference per unit length. Let ll
238 Aptitude Test Problems in Physics
be, the lengthjfof the lower part of the duct where
the cast iron is molten, and I2 be the length of
the upper part where the cast iron is in the solid phase. Then we can write the condition of
the constancy of heat flux in the form (the crosssectional area of the duct is constant)
%1lq(T1""Tme1t)= Ksol (Tmelt··· T2)
li Z2 ’
where xuq and xw] are the thermal conductivities
of the liquid and solid cast iron. Considering that
xuq = lmao], we obtain
ll: lzk(T1·——Tme1t)
T melt ‘·" T 2 •
The total length of the duct is ll —|— 1,,. Therefore,
the part of the duct occupied by the molten metal
is determined from the relation
I1 = k (T1""Tme1t)
Irl-is k(Ti—Tmeu)+(Tme1¢—T2) °
2.35*. The total amount of heat Q emitted in
space per unit time remains unchanged since it is
determined by the energy liberated during the
operation of the appliances of the station. Since
only the outer surface of the screen emits into space
(this radiation depends only on its temperature),
the temperature of the screen must be equal to
the initial temperature T = 500 K of the station.
However, the screen emits the same amount of heat
Q inwards. This radiation reaches the envelope
of the station and is absorbed by it. Therefore,
the total amount of heat supplied to the station
per unit time is the sum of the heat Q liberated
during the operation of the appliances and the
amount of heat Q absorbed by the inner surface
of the screen, i.e. is equal to 2Q. According to
the heat balance condition, the same amount of
heat must be emitted, and hence
.<L:.Z`.1
2Q T;} .
Solutions 239
where Tx is the required temperature of the envelope of the station. Finally, we obtain
2;,:.%/2r Q. ooo K.
2.36. It follows from the graph (see Fig. 67) that
during the first 50 minutes_the temperature of the
mixture does not change and is equal to 0°C.
The amount of heat received by the mixture from
the room during this time is spent to melt ice. In
50 minutes, the whole ice melts and the temperature of water begins to rise. In 10 minutes (from
1:1 = 50 min to rz = 60 min), the temperature
increases by AT = 2 °C. The heat supplied to the
water from the room during this time is q =
cwmw AT = 84 kJ. Therefore, the amount of heat
received by the mixture from the room during the
first 50 minut_es is Q = 5q = 420 kl . This amount
of heat was spent for meltingl ice of mass mice: Q =
Qtmicc. Thus, the mass of t e ice contained in the
bucket brought in the room is
2.37*. We denote by on the proportionality factor
between the power dissipated in the resistor and
the temperature difference between the resistor
and the ambient. Since the resistance of the resistor
is R1 at T3 = 80 °C, and the voltage across it is
U1, the dissipated power is Uf/R1, and we can
write
U2
—Ri-=¤¤(T3—T0)· (1)
The temperature of the resistor rises with the
applied voltage since the heat liberated by the
current increases. As the temperature becomes equal
to T1 = 100 °C, the resistance of the resistor abruptly increases twofold. The heat liberated in it will
decrease, and if the voltage is not very high, the
heat removal turns out to be more rapid than the
liberation of heat. This leads to a temperature
drop to T2 = 99 °C, which will cause an abrupt
240 Aptitude Test Problems in Physics
change in the resistance to its previous value, and
the process will be repeated. Therefore, current
oscillations caused by the jumpwise dependence
of the resistance on temperature will emerge in the
circuit.
During these oscillations, the temperature of
the resistor is nearly constant (it varies between
T2 = 99 °C and T1 = 100 °C) so that we can
assume that the heat removal is constant, and the
removed power is ot (T1 —— T0). Then, by introducing the time tl of heating (from 99 °C to 100 °C),
the time tz of cooling, and the oscillation period
T =-· t1—|—t2, we can write the heat balance equations:
U2.:
-§f·=¤¤(Ti—T¤>¢i+6'(Ti-T2),
U2: (2)
—_,§‘;-&=¤¤(Ti—T0)¢2—C(Ti—T2)·
Using the value of ot obtained from Eq. (1), we
find that
,1: C (T 1"'T2)
Ug/R1-U? (T1-· T0)/lR1(Ta"‘To)l ’
t = C (T 1-T z)
2 U%(Ti—T0)/[Ri(T3—T0)l—U%/R2°
Substituting) the numerical values of the quantities, we o tain tl = tz = 3/32 s z 0.1 s and
T cx 0.2 s.
The maximum and minimum values of the
current can easily be determined since the resistance abruptly changes from R1 = 50 Q to R2 =·-·
100 Q in the process of oscillations. Consequently,
UU
ImaX=·Ei-= 1.6 A, [mm: —é—= 0.8 A.
It should be noted that the situation described
in the problem corresponds to a f1rst—order phase
transition in the material of the resistor. As a
result of heating, the material goes over to a new
Solutions 241
phase at T1 = 100 °C (this transition can be
associated, for example, with the rearrangement
of the crystal lattice of the resistor material). The
reverse transition occurs at a lower temperature
T2-:99 °C. This phenomenon is known as hysteresis and is typical of first-order phase transitions.
2.38. Raindrops falling on the brick at first form
a film on its surface (Fig. 204). The brick has a
/
Fig. 204
porous structure, and the pores behave like capilaries. Due to surface tension, water is sucked into
pores-capillaries. The capillaries are interconnected
and have various sizes, the number of narrow capillaries _being larger. The force of su1·face tension
sucking water in wide capillaries is weaker than
the force acting in narrow capillaries. For this
reason, the water film in wide capillaries will bulge
a1nd_break. Tgiis phenomenon is responsible for the
1ss1ng soun s.
2.39. The work A done by the gas is the sum of two
components, viz. the work A1 done against the
force of atmospheric pressure and the work A2
done against the force of gravity. The mercury·gas
interface is shifted upon the complete displacement
of mercury by 2l —l— l/2 = (5/2)l, and hence
5
A1 °`—: *2* p0Sl•
The work A2 done against the force of gravity
is equal to the change in the potential energy of
mercury as a result of its displacement. The whole
of mercury rises as a result of displacement by l
relative to the horizontal part of the tube. This
16-0771
242 Aptitude Test Problems in Physics
quantity should be regarded as the iinal height of
the centre of mass of mercury. The initial position
of the centre of mass of mercury is obviously ho ==
l/8. Hence we can conclude that
L7
where M = 2lSpm,,,. is the mass of mercury.
Finally, we obtain
A——A A——5 Sl 7 Sl”~77J
- 1+ 2--*5*1% +-2*-pmgrg —- · ·
2.40. The work A done by the external force as
a result of the application and subsequent removal
of the` load is determined bg the area ABCD of the
figure (see Fig. 69}; Accor ing to the first law of
thermodynamics, t e change in the internal energy
of the rod is equal to this work (the rod is thermal y
insulated), i.e.
AW = A = ka:0 (.1: -· ::0).
On the other hand, AW = C AT, where AT is the
change in the temperature of the rod, from which
we obtain
___ AW __ kx0(a:-::0)
AT-- C -—————-C-——-.
2.41. Let the cylinder be filled with water to a
level as from the base. The change in buoyancy is
equal to the increase in the force of gravity acting
on the cylinder with water. Hence we may conclude that Ah = sz:. From the equilibrium condition for the cylinder, we can write
p25 = Pos + ms.
where pz is the gas pressure in the cylinder after
its filling with water, i.e. pz = po -l— mg/S. Using
Boyle’s law, we can write pz (h — Ah) = plh,
Solutions 243
where pl is the initial pressure of the gas. Finally,
we obtain
p = Po+ mg/S
I 1 -Ah/h ·
2.42. Let us choose the origin as shown in Fig. 205.
Then the force acting on the wedge depends only
.5/
ix
Fig. 205
on’the a:-coordinate of the shock front. The hori
zontal component of this force is
px=p0,,taua=£9£EE.= .&>.‘?.€‘Z£ ,
bb
where as = vt is the wavefront coordinate by- the
moment of time t from the beginning of propagation
of the wave through the wedge. The acceleration
imparted to the wedge at this moment of time is
_ Fx ___ pocavt
°' “"°Ez""_bxZ—°
At the moment of time to when the wavefront
reaches the rear face of the wedge, i.e. when the
wavefront coordinate is b = vto, the acceleration
of the wedge becomes .
__ p ca
an- —J’-"T- ,
1s•
244 Aptitude Test Problems in Physics
Since the acceleration of the wedge linearly depends
on time, for calculating the velocity u of the wedge
by the moment of time to we can use the mean
value of acceleration dm = poca/(2m):
p abc
u = dmlo == -5%;- ,
When the entire wedge is in the region of an elevated pressure, the resultant force acting on the
wedge is zero. The answer to the problem implies
that the condition u < v means that pn <
2mv2/(abc).
3. Electricity and Magnetism
3.1. D_ue to polarization of the insulator rod AB,
the pomt charge —|—q1 will be acted upon, in addition to the point eharge --q,, by the polarization
charges formed at the ends of t e rod (Fig. 206).
. A .............1?`
® IZ il e
qi gg
Fig. 206
The attractive force exerted by the negative charge
1ndu_ced at the end A will be stronger than the repulsive force exerted by the positive charge induced at the end B. Thus, the total force acting
on the charge —}— I will increase.
3.2. In the immediate proximity of each of point
charges, the contribution from the other charge
to the total field strength is negligibly small, and
hence the electric field lines emerge from (enter)
the_ charge in a spatially homogeneous bundle,
their total number eing proportional to the magmtude of the charge. Only a fraction of these lines
gets into a cone with an angle 2oz at the vertex
near the charge -|-qi. The ratio of the number of
these lmes to t e total number of the lines emerging
from the charge —|-ql is equal to the ratio of the
areas of the corresponding spherical segments:
2nRR (1 —c0s cz) 1
Since the electric field lines connect the two charges
of equal magnitude, the number of lines emerging
246 Aptitude Test Problems in Physics
from the charge +91 within the angle 2oz is equal
to the number of lines entering the charge ——q2
at an_angle 25. Consequently,
rail(1—¤¤¤¤)=lq2l(1—¤¤¤¤>.
whence
· B · Q 1/ l *11 l
8111 —— = I1 —— ———— ,
2 S1 2 l qzll
If 1/I q1 | / | qa | sin (ot/2) > l, an electric field
line will not enter the charge -—q2.
3.3. Before solving this roblem, let us formulate
the theorem which will be useful for solving this
and more complicated problems. Below we shall
give the proof of this theorem applicable to the
specific case considered in Problem 3.3.
If a charge is distributed with a constant density o over a part of the spherical surface of radius
R, the projection of the electric field strength due
to this charge at the centre of the spherical surface
on an arbitrary direction a is
1o
E¤=‘z:2z;Ta· Sw
where S M is the area of the projection of the part
of the surface on the plane perpendicular to the
direction a.
Let us consider a certain region of the spherical
surface ("lobule") and orient it as shown in Fig. 207,
i.e. make the symmetry plane of the lobule coincide
with the z- and x-axes. From the symmetry of
charge distribution it follows that the total field
strength at the origin of the coordinate system
(point O) will be directed a ainst the z-axis (if
0 > 0), and the field strengtilx components along
the x- and y-axes will be zero.
Let us consider a small region of the surface
AS of the lobule. The vertica component of the
Solutions 247
electric Held strength at point O produced by the
surface element AS is given by
10
AE COS (P,
where cp is the angle between the normal to the
area element and the vertical. But__AS cos cp is the
Z
S0
i as
(b
,
/ ···‘.. ·‘ :
~ ..;..-3* ’
\\ I =.——....· /
\\\\ /'I/ll
Fig. 207
area of the projection of the element AS on the
horizontal p ane. Hence the total field strength
at point O can be determined from the formula
,__ 1 oS
E — lmao T? ’
where S is the area of the hatched figure in Fig. 207,
which is the projection of the 'lobule on the horizontal plane :1:Oy. Since the area of any narrow
strip of this region (black region in Fig. 207) is
sma ler than the area of the correspon ing strip
of the large circle by a factor of sin (cz/2), the entire
248 Aptitude Test Problems in Physics
area of the hatched region is smaller than the area
of the large circle by a factor of sin (oc/2). Hence
,__ 0S __ o 2. oc__osin(oc/2)
E " 4ns0R2 “ 4m,,R¤ "R Sm 2 “ 48, ·
In the case of a hemisphere, ot = xt and
o
E ····· •
3.4. It can easily be seen from symmetry considerations that the vector of the electric field
strength Eroduced by the "lobule" with an angle oc
lies in t e planes of longitudinal and transverse
symmetry of the lobule. Let the magnitude of
this vector be E. Let us use the superposition principle and complement the lobule to a hemisphere
charged with the same charge density. For this
purpose, we append to the initial lobule another
obule with an angle at — oc. Let the magnitude
of the electric field strength vector produced by
this additional lobule at the centre of the sphere
be E'. It can easily be seen that vectors E and E'
are mutually perpendicular, and tl1eir vector sum
is equal to the electric field vector of the hemisphere at its centre. By hypothesis, this sum is
equal to E0. Since the angle between vectors E
and E0 is sr:/2 —— oc/2, we obtain
E = E0 sin -;- ,
3.5*. Let us consider the case when the capacitors
arezoriented so that the plates with like charges
··Ig|,+ ·Z`+Z -+-Iil··*—!——-—>l
Fig. 208
Iace each other (Fig. 208). The field produced by
the iirst capacitor on the axis at a distance x
Solutions 249
from the positive plate is
E (xl- [mso [:::2 (z:-{-l)“ il" 4:re0x3 (3: > l)°
The force _acting on the second capacitor situated
at a distance d from the first is
=. E a .. =.<!z.‘!i [.L.__L.]
F qzl () E(d-I-D] 4,,80 d3 (d_H)3
N 3q1Q_£__
N 2ne,,d·* °
Therefore, the capacitors will repel each other in
this case.
A similar analysis can be carried out for the
case when the capacitors are oriented so that the
plates with unlike charges face each other. Then
the capacitors will attract each other with the
same force
_ 3 q1q•zl2
F -72.- :rts0d* '
3.6. We choose two small arbitrary elements belonging to the surfaces of the first and second hemi..sphe1·es and having the areas AS1 and AS2. Let the
separation between the two elements be rm. The
force of interaction between the two elements can
be determined from Coulomb’s law:
AF12 - ZE @-0] AS1 U2 U1O`g.
In order to determine the total force of interaction between the hemispheres, we must, proceeding
from the superposition principle, sum up the forces
AFI, for all t e interacting pairs of elements so
that the resultant force of interaction between the
hemispheres is
F = kU1G2,
where the coefiicieut k is determined only by the
geometry of the charge distribution an by the
250 Aptitude Test Problems in Physics
choice of the system of units. If the hemispheres
were charged with the same surface density 0,
the corresponding force of interaction between the
hem-ispheres would be
F = kG2,
where the coefficient k is the same as in the previous
formula.
Let us determine the force F. For this purpose,
we consider the "upper" hemisphere. Its small
surface element of area AS carries a charge Aq =
0 AS and experiences the action of the electric
iield whose strength E' is equal to half the electric
field strength produced by the sphere having a
radius R and uniformly charged with the surface
density 0. (We must exclude the part produced by
the charge Aq itself from the electric field strength.)
The force acting on the charge Aq is
,__ 1 4atR’ ___ oz ,
end is directed along the normal to the surface
element. In order to find the force acting on the
upper hemisphere, it should be noted that according
to the expression for the force AF the hemisphere
as if experiences the action of an effective pressure
p = 0“/(212,,). Hence the resultant force acting on
the upper hemisphere is
17-- :rtR2 —— fi ¤R=
_P `— Zeo
(although not only the "lower" hemisplhere, but
all the elements of the "upper" hemisp ere make
a contribution to the expression AF for the force
acting on the surface element AS, the forces of
interaction between the elements of the upper hemisphere will be cancelled out _in the general expression for the force of interaction between the hemispheres obtained above).
Since F = k0*, we obtain the following} expression for the force of interaction between t e hemi-
Solutions 251
spheres for the case when they have different surface charge densities:
R2
F"—= koloz =*E··g)— 010,2.
3.7. The density of charges induced on the sphere
is proportional to the electric field strength:
o cx E. The force acting on the hemispheres is
proportional to the field strength:
F 0C oSE 0c R“E“,
where S is the area of the hemisphere, and R is its
radius. As the radius of the sphere changes by a
factor of n, and the field strength by a factor of k,
the force will change by a factor of k”n”. Since the
thickness of the sphere walls remains unchanged,
the force tearing the sphere per unit length must
remain unchanged, i.e. kznz/n = 1 and k =
1/ }/n = 1/ 1/ 2. Consequently, the minimum electric field strength. capable of tearing the conducting
shell of twice as large radius is
E
E =——°_
1•
3.8. Let 1 be the distance from the largle conducting
sphere to each of the small balls, d t e separation
between the balls, and r the radius of each ball.
If we connect the large sphere to the first ball,
their potentials become equal:
1 Q Q1 )__
‘¤;?s;( z at T ·‘*’· “’
Here Q is the charge of the large sphere, and qa is its
potential. If the large sphere is connected to the
second ball, we obtain a similar equation correspondinglto the equality of the potentials of the
large sp ere and the second ball:
1 Q *11 qa )__
4m,(z+d+. ·‘*’· (2)
252 Aptitude Test Problems in Physics
(We assume that the charge and the potential of
the large sphere change insignificantly in each
charging of the balls.) When the large sphere is
connected to the third ball, the first and second
balls being charged, the equation describing the
equality of potentials has the form
1 Q qi q2 qa )__
aw, ( i +Tz"+7i`+ r ··‘*’· (3)
The charge qs can be found by solving the system
of equations (i)—(3):
q3 •:.£.
(11 °
3.9. The charge q1 of the sphere can be determined
from the formula
q1 = 4ne0cp1r1.
After the connection of the sphere to the envelope,
the entire charge q1 will flow from the sphere to the
envelope and will he distributed uniformly ove1·
its· su1·face. Its potential (pz (coinciding with the
new value of the potential of the sphere) will be
-....41 ,_., Li.
(P2- Zutsorz (P1 r2 '
3.10. We shall write the condition of the equality
to zero of the potential of the sphere, and hence
of any point inside it (in particular, its centre), by
the moment of time t.`We shall single out three
time intervals:
aabb
(i) ¢<—;» (2) ;—<¢<—;-. (3) ¢>—;—.
Denoting the charge of the sphere hy q (t), we obtain
the following expression for an instant t from the
first time interval:
qi _q;_ q(¢) _
a + b + vt _0’
Solutions 25 3
whence
q(¢)=-—v(-$f—+-%f-) ¢.
... - 3.; 3;.
aw- »( G + , ).
For an instant t from the second time interval, we
find that the fields inside and outside the sphere
are independent, and hence
q (0+ qi __ _;1_q_ _ qs
"_*—v¢ r b· ’=<"·*"T·
Finally, as soon as the sphere absorbs the two
point charges q1 and qs, the current will stop flowing
through the "earthing" conductor, and we- can write
I , (t) = O.
Thus,
... 3;. Q. L
q - .. ss .1 J;
IU)- vb. v<¢<v»
b
01 t > 7 •
3.11. Taking into account the relation between
the capacitance, voltage, and charge of a capacitor,
we can write the following equations for the three
capacitors:
- E.; .. =.‘b. .. =.‘!L
(PA (P0 C1, (PB (P0 C2» (PD (P0 C8»
where C1, 0,, and C, are the capacitances of the
corresponding capacitors, and q , q,, and q are
the charges on their plates. According to the charge
conservation law, q1 —l— qq + q, = 0, and hence
the potential of the common point 0 is
(P = <PAC1+<PBC¤+*Pr>Ca
° 61+ 63+ Cs '
254 Aptitude Test Problems in Physics
3.12*. Since the sheet is metallic, the charges
must be redistributed 0VB1° its surface so that the
field in the bulk of the sheet is zero. In the first
approximation, we can assume that this distribution is uniform and has density -—o and 0 over
the upper and the lower surface respectively of the
sheet. According to the superposition principle,
we obtain the condition for the absence of the field
in the bulk of the sheet:
QU_
4:rte0l’ so _0'
Let us now take into consideration_ the'nonuniformity of the field produced _by the point charge
since it is the single cause of the force F of interaction. The upper surface of the sheet must be attracted with af0rce0Sq/(4:1:-:,,12),whilethe lowersurface
must he repelled with a force oSq/[4ne0(l—l-d)“].
Consequently, the force of attraction of the
sheet to the charge is
F: JQ. [1.....L_ N <1”$d
4ns0l* (1-{-d/l)° il" 8n’e0l‘• °
3.13. It can easily be seenithat the circuit diagram
proposed in the problem is a "regula1·" tetrahedron
‘/®
0
Fig. 209
Solutions 255
whose edges contain six identical capacitors.
Therefore, we conclude from the symmetry considerations that irrespective of the pair of points
between which the current source is connected,
there always exists an uncharged capacitor in the
circuit (the capacitor in the edge crossed with
the edge containing the source). For example, if
the current source is connected between points A
and B in Fig. 2.09, the cadpacitor between points C
and D will be uncharge since the potentials of
points C and D are equal.
3.14. The capacitance of the nonlinear capacitor is
C = eC0 = otUC,,,
where C0 is the capacitance of the capacitor without
a dielectric. The charge on the nonlinear capacitor
is qu = CU = otC0U“, while the charge on the
normal capacitor is q = C, U. It follows from the
charge conservation law
qi! + q0 L: Co Un
that the required voltage is
U = .l./é2*.&j.:L·i = 12 v.
2oz
3.15. Let us go over to the inertial reference frame
hxed to the moving centre of the thread. Then
the balls have the same velocity v at the initial
instant. The energy stored initially in the system is
__ q2 2mv2
W1"' 4:te0·2l + 2 ·
At the moment of the closest approach, the energy
of the system is
.. A;
W2- 4:n:e,,d °
Using the energy conservation law, we iind that
d_ 2lq“
_ q*-|-8:rtc0inv’l °
256 Aptitude Test Problems in Physics
3.16. Let v1 and vg be the velocities of the first and
second balls after the removal of the uniform electrio field. By hypothesis, the angle between the
velocity vi and the initial velocity v is 60°. Therefore, the change in the momentum of the first ball is
Ap, = q1E At == mlv sin 60°.
Here we use the condition that vi = v/2, which
implies that the change in the momentum Apl
of the first ball occurs in a direction perpendicular
tojthe direction of its velocity v1.
{Since E || Ap, and the direction of variation
of ,_the second ball momentum is parallel to the
direction of Apl, we obtain for the velocity of the
second ball (it can easily be seen that the charges
on the balls have the same ·sign)
v =» mi soo:-F:
2•
The corresponding change in the momentum
of the second ball is
Apr-q¤E Af- COS 30. .
Hence we obtain
rn - .;~.isi2§2°; .2;.... .’; an - .4. ,,
qa —_ 300 , Ina -—- 3 mul C- 3 1.
3.17. The kinetic energy of the first ball released
at infinity (after a long time) can be determined
from the energy conservation law:
mv? _ q2 ( 1 1 _ 1
2 "4zre0 a1+a2+° ° '+aN_1)’
where al, a , . . ., aN_1 are the distances from
the first ball (before it was released) to the remaining balls in the circle, al and aN-1 being the
distances to the nearest neighbours, i.e. al =
aN,1°;a -'=
Solutions 257
Analyzing the motion of the second ball, we
neglect the influence of the first released ball.
Then
mug _ qa 1 1 1 )
i.e. one of the nearest neighbours is missing in the
parentheses. Therefore,
K=..~123.....~&=.ai_
2 2 43'|Z80G °
or
3.18. According to the momentum conservation law,
mv = (m —l- M) u,
where m is the mass of the accelerated particle,
M is the mass of the atom, and u is their velocity
immediately after the collision.
We denote by Wim, the ionization energy and
write the energy conservation law in the form,
22
.,2..;.. : Wim + _
Eliminating the velocity u from these equations,
we obtain
——;,‘i—=w....(¤+T,—).
If m is the electron mass, then m/M < 1, and the
kinetic energy required for the ionization is
s
*'%L" N Wi0n•
When an atom collides with an ion of mass
m as M, mvz/2 as ZWIOD, i.e. the energy of the
ion required for the ionization must be twice as
high as the energy of the electron.
17-0771
258 Aptitude Test Problems in Physics
3.19. From the equality of the electric and elastic
forces acting on the free ball,
q' _
4:rte0·4l“ "`kl
we obtain the following expression for the length Z
of the unstretched spring:
ls=...Q3;...
16:wok ’
where k is the rigidity of the spring, and q are the
charges of the alls.
Let us suppose that the free ball is deflected
from the equilibrium position by a distance .1:
which is small in comparison with l. The potential
energy of the system depends on as as follows:
W (x)=i— k (l-—x)2-|—-——Q——— z 2 k12-{-kx¤
Here we have taken into account the relation between ql, k, and Z obtained above and retain the
terms of the order of [:1:/(2l)]“ in the expression
...L.........1..._
2l-as — 2l (1-x/2l)
1 :c .1: 2
··‘wl‘+(w)+(‘.·a") +·
Thus, it is as if the stretched spring has double the
rigidity, and the ratio of the frequencies of harmonic vibrations of the system is
2.;;......l/E .. yi-;
V1 •
3.20. Let us consider the ·two charged balls to be
a single mechanical system. The Coulomb interaction between the balls is internal, and hence it
does not affect the motion of the centre of mass.
The only external force acting on the system is
the force of gravity. It is only this force that will
Solutions 259
determine the motion of the centre of mass of the
system. Since the masses of the balls are equal,
t e initial position of the centre of mass is at a
height (hl -l- hz)/2, and its initial velocity v is
horizontal. Then the centre of mass will move
along a parabola characterized by the following
equation:
_ ki-thi g ¤¤ 2
h·—2—-(all?) "’
where a: is the horizontal coordinate of the centre
of mass, and h is its vertical coordinate. At the
moment the first ball touches the ground at a
distance x = l, the height H of the centre of mass,
according to expression (1), is
___ h1+hz 8 I )2
H—2(2v
Since the masses of the balls are equal, the second
ball must be at a height H2 = 2H at this instant.
Therefore,
g2
Ha=h1+ha···8(·;·)
3.21. Let the resistance of half the turn be R.
Then in the former case, we have fifteen resistors
of resistance R connected in parallel, the total
resistance being R/15.
In the latter case, we have the same fifteen resistors connected in series, the total resistance
being 15R. Therefore, as a result of unwinding,
the resistance of•the wire will increase by a factor
o 225.
3.22. It can easily be noted from symmetry considerations that the potentials of points A and C
(Fig. 210) at any instant of time will be the same.
Therefore, the closure of the key K will not lead
to any change in the operation of the circuit, and
the coil AC will not be heated. •
17•
260 Aptitude Test Problems in Physics
3.23. After adding} two conductors, the circuit will
acquire the form s own in Fig. 211. From symmetry
considerations, we conclude that the central conFig. 210
ductor will not particigate in electric charge
transfer. Therefore, if t e initial resistance R1
of the circuit was 5r, where r is the resistance of
Fig. 211
a conductor, the new resistance of- the circuit will
become
2r
.R2=2T •"= 3T.
Therefore,
R1 —— 5 ·
3.24. The total resistance RAB of the frame can
easily be calculated by noting from symmetry
considerations that there is no current in the edge
Solutions 261
CD: RAB -= R/2, where R is the resistance of
an edge. Therefore,
2U
I‘T*
where U is the applied voltage.
The total current can be changed in two ways:
(1) if we remove one of the edges AD, AC, BC or
BD, the change in the current will be the same;
(2) if we remove the edge AB, the change will be
different. In the first case, the change in the current
will be AI = -—(2/5) U/R = —I/5, and in the
second case, the total resistance will be R, and
hence AI = —U/R = —I/2 = Alma,.
3.25. It follows from symmetry considerations
that the potentials of points C and D are equal.
Therefore, this circuit can be replaced by an equivalent one (we combine the junctions C and D,
GM)
·A__{5
lx
: l (/3,)
. .. .2
\\L:J///
+
rw
Fig. 212
Fig. 212). The resistance between points A and B
of the circuit can be determined from the formulas
for parallel and series connection of conductors:
_ R/2(H/2—|—R) _ 3 R __§_
R°"D"’°°" R/2—|—R/2-}-R `“7C 2 8 R’
262 Aptitude Test Probl_ems in Physics
R37
RAc<n)B=j—+·B·R=·8·H»
_ R (7/8) R _ 7
RAB" —R-—l—(7/8)R " is R‘
Thus, the current I in the leads can be determined
from the·—__iormula
1- ...’!........E. IL
" (7/15)R"` 7 R °
3.26*. In order to simplify the solution, we present
the circuit in a more symmetric form (Fig. 213).
The obtained circuit cannot be simplified by connecting or disconnecting junctions (or by removing
some conductors) so as to obtain parallel- or seriesconnected subcircuits. However, any problem
involving a direct current has a single solution,
which we shall try to "guess" by using the symmetry of the circuit and the similarity of the currents in the circuit.
Let us apply a voltage U to the circuit and mark
currents through each element of the circuit. We
U 0___
Fig. 213
shall require not nine values of current (as in the
case of arbitrary resistanoes of circuit elements)
but only tive values I1, 1,, I , I , and I5
(Fig. 214). For such currents, Kircl1hoH"shrst rule
for the junction C
I1 '·= Is + In
Solutions 263
and for the junction D
Is + !• = T4 + It
will automatically be observed for the junctions E
and F (this is due to the equality of the resistances
a
I [5 F
Fig. 2
of all resistors of the circuit). Let us now write
Kircbhoffs second rule in order to obtain a system
of tive independent equations:
(Y¤+Is+}1)R -‘= U,
(I, + I4) R :` ISR,
(Il + Is) R = IJ?.
where R is the resistance of each resistor. Solvini
this system of tive equations, we shall express al
the currents in terms of I1:
6I3
Is:-"g‘I1• Is=*§"• I4="5*I1»
I·=*g' I1•
Besides,
46
*5 R.
Consequently, U/I1 =R.
264 Aptitude Test Problems in Physics
Considering that the resistance R A B of the
circuit satisfies the equation R A B = U/(I1 —l— I 2),
we obtain
R B .. i..........9..........§.. J!.
A Ii+I2 Irl-(6/5)!i ii I1'
Taking into account the relation obtained above,
we get the following expression for the required
resistance:
15
RAB —· T R.
3.27. It follows from symmetry considerations
that the initial circuit can be replaced by an equivalent one (Fig. 215). We replace the “inner"
R/z g e/2
1
R/ /2
,4 1 5
Fig. 215
triangle consisting of an infinite number of elements by a resistor of resistance RA B/2, where the
resistance RA B is such that RA B ·-= Rx and RAB =
ap. After simplification, the circuit becomes a system of series- and parallel-connected conductors.
In order to {ind Rx, we write the equation
____ RR,,/2 RRx/2 )—i
RFB (R+ R+Rxx2) (R+R+ Ha-Hx/2 ·
Solutions 265
Solving this equation, we obtain
12 7-1 a 7·-1
33
3.28. It follows from symmetry consideraions
that if we remove the iirst element from the circuit,
the resistance oi the remaining circuit between
points C and D will be RC = kRAB. Therefore,
the equivalent circuit of the infinite chain will
2
A6.
#2 ka.
0 12
Fig. 216
have the form shown in Fig; 216. Applying to this
circuit the formulas for t _e resistance o seriesand parallel-connected resistors, we obtain
R —I—R RCRAB
3 = .;...1...... _
AB Ba·|·kR,4B
Solving the quadratic equation for RAB, we obtain
(in particular, for k = 1/2)
R B: Ri—R2·|· l/R i-|·B5·|·6R1R
A2°
3.29. The potentiometer with the load is equivalent
to a resistor of resistance
_ R RR/2 ___ 5°
R1 · *7 +‘aT1in‘ · Ti RHence the total current in the circuit will be
1 —...L. --i`L E.
*" (5/6);: " 5 1z·
266 Aptitude Test Problems in Physics
The voltage across the load will be
U1]=U-—I1
If the resistance of the load becomes equal to 2R,
the total current will be
1 —........H..........-J£ IL
2* _1i_+ (R/2)(2R_2_ — 9 R ’
2 R/2—|—2R
The voltage across the load will become
Uz1=U—I2 -gi:-3U
Thus, the voltage across the load will change by
a factor of k = U2;/U1]:
__ U2] _ 10 .
k‘_ Ull __. 9
3.30. In the former case, the condition I1 == I,
is fulfilled, where I1 = mln, and I 2 = mznz. Consequently, mln, = m,n,. In the latter case, I QR, ==
IgRx, where Rx is the resistance of the second
resistor. Besides, I; = mln; and Ig = mzng, and
hence
Rldlfli == Rxa2n;•
Finally, we obtain
..@.’5.=....R=·=”$
nl nz
Therefore,
Rx = .§.L'2;l. _
nin,
3.3l. The condition required for heating and melting the wire is that the amount of Joule heat
liberated in the Jwrocess must be largter than the
amount of heat issipated to the am ient:
FR > kS (T — Tam).
l Solutions 267
By hypothesis, the current required for melting
the first wire must exceed 10 A. Therefore,
'=·4dJ (Tmelt ··- Tam) = I%F1»
where I is the length of the wires, Tmclt is the melting point of the wire material, and I1 and R, are
the current and resistance of the first wire.
The resistance of the second wire is R, =
R1/16. Therefore, the current I , required for melting
the second wire must satisfy the relation
IER2 > k•’*dal (Tmeu ·· Tam)Finally, we obtain
I 2 > 81, = 80 A.
3.32. Let the emf of the second source be ‘8,.Theu,
by hypothesis,
I: $1+% :__§;
where R is the resistance of the varying resistor
for a constant current. Hence we obtain the answer:
$5
Rx: TL
3.33. The current through the circuit before the
source of emf Z, is short-circuited satisfies the
condition
11:: 81+xs
R+n+¤’
where r, and r, are the- internal resistances of the
current sources. After the short-circuiting of the
second source of emf 32, the current through the
resistor of resistance R can be determined from the
formula
.. $1
’=·2¤z¢;· ·
268 Aptitude Test Problems in Physics
Obviously, if
$1 > $z+$1
R+"1 R+"1+"z ,
he answer will be affirmative.
Thus, if the inequality 51 (R —|— rl + rz) >
(51+ 52) (R + rl) is satisfied, and hence
51r, > 5, (R —l- rl); the current in the circuit
increases. If, on t econtrary 5 r < 5 (R —}— r),
tbz short-circuigng of the currtehg sourde leads lo
a ecrease in t e current in t e circuit.
3.34*. Let us make use of the fact that any "black
box" circuit consisting of resistors can be reduced
7 {` ``—`````````` T
5
I P2 2
•PP1
2I4{sl4
s. .... .. ........ J
Fig. 217
ti? a ferr; (Fig. 217), wghere the quémtiities R1,
, . ., are expresse in terms o t e resist·
arilces of the initial resistors of the "black box"
circuit (this can be verified by using in the initial
cirguit the transformations of the star-delta type
an reverse transformations). By hypothesis,
ecgual currents pasis each tiime through resistors
o resistance R an R an also throu h R and
R5 (or there islno current through then? wheli the
clamps are disconnected). Using this circumstance,
av; canhsirlripliféithe circuit as shown in Fig. 218.
€¤» Y YP0 9518,
U? U2
N =···r·····:· » N =··7·····v···r···:·········
I Rrl·R¤ ” R. —l-1*2}*3/<R2+R§)
U* U2
' 1*2+1% ’ " Rs-l-R13;/(R;+R;)
Solutions 269
It `can easily be verified that
N1N,==N2N3.
Consequently,
3.35. At the first moment alter the connection of
the key K, the capacitors are not charged, and
I" '“''“`““““`` ""I
,,’5
: R, pr P, {
I2I
II
L...... _..... ....-......J
Fig. 218
hence the voltage between points A and B is zero
(see Fig. 93). The current in the circuit at this
instant can be determined from the condition
8I
I =*··*** ,
1 R1
Under steady-state conditions, the current between pointsA and B will pass through the resistors R1 and R3. Therefore, the current passing
through these conductors after a long time is
I2 T-""""'$;_""' •
R1`l'R3
3.36. Let us consider the steady-state conditions
when the voltage across the capacitor practically
does not change and is equal on the average to UB,.
270 Aptitude Test Problems in Physics
When the key is switched to position 1, the
charge on the capacitor will change during a short
time interval At by
At ($1‘*Ust)
R1 •
When the key is switched to position 2, the charge
will change by
At ($z"‘Ust)
R2 ·
During cycle, the total change in charge must be
zero:
($1 ··· Ust) (lg: ·· U st)
·-—————- ·—-———————-=O.
R1 {_ R2
Hence the voltage Ust and the capacitor charge qu
in the steady state can be found from the formulas
$2R1+$1R2
sy _.=____________
L st R1+Rz ,
C ($2Ri+ $$1}%)
Z LL" •
qSt Bt R1+Ra
3.37. A direct current cannot pass through the
capacitors of capacitance C1 and C . Therefore,
in the steady state, the current through the current
source is
__ ii
’ · TimiSince the capacitors are connected in series, their
charges q will be equal, and
..2. L:
Consequently,
q__ ?5R2C1C2
("+H2) (C1'} C2) •
Solutions 271
The voltages U and U, across the capacitors can
be calculated irom the formulas
U1: .2.:..._..._.l$R=C¤
Cr 0+}*2) (C';-+-Ca) ’
U ___ q $R2C1 .
a—— —··· = ———————-—
Ca (T-!-Ri) (01+ Cz)
respectively.
3.38*. We shall mentally connect in series two perfect (having zero internal resistance) current sources
of emf’s equal to —U,, and UO between points A
and F. Obviously, this will not introduce any
change in the circuit. The dependence of the current
through the resistor of resistance R on the emf’s of
the sources will have the form
I = 0:,5 -— BU0 —l- BU0,
where SS is the emf of the source contained in the
circuit, and the coefficients ot and B depend on the
resistance of the circuit.
If we connect only one perfect source of emf
equal to —-UO between A and F, the potential
difference between A and °B will become zero.
Therefore, the first two terms in-the.equation
for I will be compensated: I = BU 0. The coefficient
B is obviously equal to 1/(R —|— Rc"), where Rm
is the resistance between A and B when the resistor R is disconnected. This formula is also valid
for the case R = 0, which corresponds to the connection of the ammeter. In this case,
U
I
0 Ref!
Consequently, the required current is
{ :._ ..y.*L{.•L.
Rid-UO
3.39. When the key K is closed, the voltage
across the capacitor is maintained constant and
equal to the emf *25 of the battery. Let the
displacement of the plate B upon the attainment
272 Aptitude Test Problems in Physics
of the new equilibrium position be -.1:1. In this
case, the charge on the capacitor is qi = C15 =
:10525/(d —-s x,),where S is the area of the capacitor plates. The lield strength in the capacitor is
E = ‘6/(d — as ), but it is produced by two plates.
Therefore, the tleld strength produced bit; one plate
is E1/2, and for the force acting on t e plate B
we can write
Elql __ Bglgga ___
2 ’“ 2<¢—·=l>· “""" “’
where k is the rigidity of the spring;
Let us now consider the case w en the key K
is closed for a short time. The capacitor acquires
a charge q2, = e0S$/d (the plates have no time
to shift), which remains unchanged. Let the displacement of the plate B in the new equilibrium
position be :::2. Then the field strengthn the capas,,S/(d ——- ::2). In this case, the equilibrium condition for the plate Bcan be written in the.form
Em __ qi __ ¤¤·$'iS“ ___ .
"`;'2""` zcos “ 2dz ""‘*·‘ (Z)
Dividing Eqs. (1) and (2) termwise, we obtain
4;, =-= rr, [(d -— 4::,)/dlz. Considering that xl == 0.id
by hypothesis, we get
$2:-7-•0•08d•
3.40. Let us represent the central junction of wires
in the foun of two junctions connected by the wire
3v4
,l ',
Fig. 219
Solutions 273
5-6 as shown in Fig. 219. Then it follows from symmetry considerations that there is no current
through this wire. Therefore, the central junction
can be removed from the initial circuit, and we
arrive at the circuit shown in Fig. 220. By hypothesis,
R12=R13=R34 ZR24 = V,
R15=R25=R38=R46“-: %
Let U be the voltage between points 1 and 2.
Then the amount of heat liberated in the conductor
1-2 per unit time is
Us
Q1a="‘,T‘ ·
§{ From Ohm’s law, we obtain the current through
Fig. 220
the conductor 3-4:
r (1/2+ 3) '
The amount of heat liberated in the conductor 3-4
per unit time is
U2
Q =-' I2 T ; Y-:· —
18-0771
276 Aptitude Test Problems in Physics
The voltage across each of the resistors R1 and
R, can be determined from the formula
{SBIR,
U ==°°—-I =—-———————-———
R= 6 r rRi+R2 (r-PF1)
The power liberated in the resistor R2 can be calculated from the formula
R’ R2 — rR1+R2(R1+r) R2
(TRIP/R2'+'2rR1(r"l`R1)'l"(r"l`R1)2 Ra °
The maximum lpower will correspond to the minimum value of t e denominator. Using! the classical
inequality az —}— bz ,> 2ab, we find t at for R =
rR1/(r -\— R1) the denominator of the fraction zhas
the minimum value, and the power liberated in
the resistor R2 attains the maximum value.
3.45. At the moment when the current througlh
the resistor attains the value I 0, the charge on t e
capacitor of capacitance C1 is
Q1 = C1I0R•
The energy stored in the capacitor by this moment is
.. ...2;.
W1 “` 201
After disconnecting the key, at the end of recharging, the total charge on the caplacitors is q1,
and the voltages across the plates of t e two capacitors are equal. Let us write these conditions in
the form of the following two equations:
»· »= .2; :.2é.
where q; and gg are the charges on the capacitors
after recharging. This gives
q, = qi6'i qi: <1i6'¤
i 01+cs , C1+C• °
Solutions 272
The total energy of the system after recharging is
rz rz 2
_ qi qa = qi
W2 "- 201 + 2C, 2 (C1—}—C2)
The amount of heat liberated in the resistor during
this time is
(IJ?)2 Cz
=W —W = —-—————-—
Q 1 “ 2 <c1+c.>
3.46. Before switching the key K, no current flows
through the resistor of resistance R, and the charge
on the capacitor of capacitance C2 can be determined from the formu a
q = $02,
The energy stored in this capacitor is found from
the formula
?S’C
WC2 Z *72*
After switching the key K, the charge q is redistributed between ·two capacitors so that the
charge ql on the capacitor of capacitance C1 and
the charge q2 on the capacitor of capacitance C2
can be calculated from the formulas
- .&.;.‘£L
<11—!—<12—q. C1 C2 .
The total energy of the two capacitors will be
w = ..£.....-...§i(£..
°*`*‘°* 2(6'i-I-Cz) 2(€'i+€'·z) '
Therefore, the amount of beat Q liberated in the
resistor can be obtained from the relation
2 2 63+02 2(6’i·|·6'2) °
3.47. By the moment when the voltage across the
capacitor has become U, the charge q has passed
278 Aptitude Test Problems in Physics
through the current. source. Obviously, q/U = C.
From the energy conservation law, we obtain
___ Q+q’
@1* 2C •
where Q is the amount of heat liberated in both
resistors. Since they are connected in parallel,
Q1/Q, = R,/R1, whence
_ U2 R1
°=·· C (i" r T) HTR:
= -——————— 225-——U .
2<H.+R.> ‘ ’
3.48. D.uring the motion of the jumger, the magnetic flux through the circuit formed y the jumper, " rails, and the resistor changes. An emf is
induced in the circuit, and a current is generated.
As a result of the action of the magnetic held on
the current in the jumper, the latter will be decelerated.
Let us determine the decelerating force F.
Let the velocity of the jumper at a certain instant
be v. During a short time interval At, the jumper
is shifted along the rails by a small distance Ax ·-·=
v At. The change in the area embraced by the
circuit is vd At, and the magnetic flux varies by
Ad) = Bvd At during this time. The emf induced
in the circuit is
Aib
g- ······K·i··— ····BUd.
According to Ohm’s law, the current through the
jumper is I = QS/R. The force exerted hy the magnetic held on the jumper is
B”d*v
According` to Lenz‘s law, the force F is directed
against t e velocity v of the jumper.
Solutions 279
Let us now write the equation ol motion for
the jumper (over a small distance Ax):
md: F = —— ,
Considering that a = Av/At and v = Ax/At, we
obtain
m Av=_ Bitzi; Ax
It can be seen that the change in the velocity of
the jumper is proportional to the change in its
a:·coordinate {at the initial instant, :::0 = 0). Therefore, the tota change in velocity Ur = vo = O —
vo = —·-vu is connected with the change in the
coordinate (with the total displacement s) through
the relation "
B2d2
m ("‘vo) = - .
Hence we can determine the length of the path
covered by the jumper before it comes to rest:
_ mRv0
° " Bad:
When the direction of the magnetic induction
B forms an angle on with the normal to the plane
of the rails, we obtain
__ mRv,,
‘ " B’d* cos*<z °
Indeed, the induced emf, and hence the current
through the jumper, is determined by the magnetic
flux -t rough the circuit, and in this case, the flux
is determined by the projection of the magnetic
induction B on the normal to the plane of the
circuit.
3.49. The lines of the magnetic flux produced by
the falling charged ball lie in the horizontal plane.
Therefore, the magnetic flux d> B through the sur-
280 Aptitude Test Problems in Physics
face area bounded by the loop is zero at any instant
of time. Therefore, the galvanometer will indicate
zero.
3.50. Let us choose the coordinate system ::0y
with the origin coinciding with the instantaneous
*P
.£/
. IB
x. ¤I
N’___ · ___
0 I `*
`___ __,,/ ·'”
mg
Fig. 222
position of the ball (Fig. 222). The a:—aXis is "centripetal", while the y-axis is vertical just as the
magnetic induction B.
The system of equations describing the motion
of the ball (we assume that the ball moves in a
circle counterclockwise) will be written in the
form
.NSina·—qvB= ,
l smc:
N cos cc=mg.
Besides,
-2-I-E;-=T, r==Zsincz.
v
Solving this system of equations, we obtain
T: l/ l*—(T/2n)*-ml ""
` [2¤/(5T) :1:.qB/(mz)l‘ '
Solutions 281
The plus sign corresponds to the counterclockwise
rotation of the ball, and the minus sign to the
clockwise rotation (if we view from the top).
3.51. When the metal ball moves in the magnetic
held, the free electrons are distributed over the
surface of the ball due to the action of the Lorentz
force so that the resultant electric held in the bulk
of the ball is uniform and compensates the action
ofthe magnetic held. After the equilibrium has
been attained, the motion of electrons in the bulk
of the metal ceases. Therefore, the electric held
strength is
Erase + qlv >< Bl = 0.
whence
Ewa = [B X v].
We arrive at the conclusion that the uniform electric held emerging in the hulk of the ball has the
magnitude
|E,eS|=|B||v|sinoz.
The maximum potential difference Aqumax
emerging between the points on the ball diameter
paral el to the vector EWS is
Acpmax = | EMS |·2r = | B | I v | sin oz·2r.
3.52. The magnetic induction of the solenoid is
directed along its axis. Therefore, the Lorentz
force acting on the electron at any instant of time
will lie in the plane perpendicular to the solenoid
axis. Since the electron velocity at the initial
moment is directed at right angles to the solenoid
axis, the electron trajectory will lie in the plane
perpendicular to the solenoid axis. The Lorentz
orce can be found from the formula F == evB.
The trajectory of the electron in the solenoid
is an arc of the circle whose radius can be determined from the relation evB = mvz/r, whence
,.. L"!.
_ eB '
282 Aptitude Test Problems in Physics
The trajectory of the electron is shown in Fig. 223
(top view), where O, is the centre of the arc AC
described by the electron, v' is the velocity at
which the electron leaves the solenoid. The segments OA and OC are tangents to the electron
I
Q
v
$ —~·
7
Fig. 223
trajectory at points A and C. The angle between
v and v{· is obviously qa = LA OIC since LOA O1·=
In order _to {ind qw, let us consider the right
triangle OA O1? side OA = R and side AO, -= r.
Therefore, tan (cp/2) = R/r = eBR/(mv). Therefore,
<p=2arctan (ig;-),
Obviously, the magnitude of the velocity remains unchanged over the entire trajectory since
the Lorentz force is perpendicular to the velocity
at any instant. Therefore, the transit time of electron in the solenoid can he determined from the
relation
... 19; .. :22... Er. .:92*..)
L" v —eB—eB &rcmn( mv ‘
Solutions 283
3.53. During the motion of the jumper, the magnetic flux across the contour "closed" by the jumper
varies. As a result, .an emf is induced in the contour.
During a short time interval over which the
velocity v of the jumper can be treated as constant,
the instantaneous value of the induced emf is
ACD
QS: --52--. —-—vbB cos cz.
The current through the jumper at this instant is
Aq
I " TSP
where Aq is the charge stored in the capacitor during
the time At, i.e.
Aq = C N6 = CbB Avcosor.
(since the resistance of the guides and the jumper
is zero, the instantaneous value of the voltage
across the capacitor is ‘6). Therefore,
1 Z cbs (-2;) cos cr: com cos e,
where a is the acceleration of the jumper.
The jumper is acted upon by the force of gravity
and Ampére’s force. Let us write the equation of
motion for the jumper:
ma = mg since -— IbB cosa = mg sinoz
—Cb”B”a cosz oz.
Hence we obtain
a __ mg sin on
— m-{—Cb”B“ cos? cc °
The time during which the jumper reaches the
foot of the "hump" can be determined from the condition t = at’/2:
_ YZ-__ VI 2l , ,
t... 1/-Z--. mgsina (m—|—Cb Bicos ca).
r284 Aptitude Test Problems in PhySi0S
The velocity of thé jumper at the be
_at__l/ Zlmgsinct
vt _ ` m—l—Cb2B2 coszot °
3.54*. The magnetic flux ooross the surface bI0I?5¤€(;%d
by th€ SUp€1‘COI1dl1ctiI1g loop is c0DS_t&Ut· R Z 0)’
AQ)/it = 3, but <6 = [R Z o (since *
and ence (D = const.
The magnetic flux through the surface boulngtii
by the l00p is the sum of the external glaégcd by
flux and the flux of the magnetic field Plghu mime
the current I assing through the loop- _ Qtam ig
the magnetic Eux across the loopjit any ms
(D ·= a2B0 —l— (12GZ —|— LI,
· . . r- 0 Eilld
Since (D ,= B az at the initial moment (Z 7.
I = 0), the ciurent I at any other instant will be
determined by the relation
otzaz
LI=—0cza°, I:—T_ ·
The resultant force errorted by th? t¥l;3i3;g§§;
field on the current loo is the sum 0 amugl
acting on the sides of tiie loop which are p
to the y—axis, i.e.
F=2a|ocx|I=a2otI
and is directed along the z-axis. _ s
Therefore, the equation of motion for the loop
has the form
O. 2
WUI; *mg+a2¤·I? •
, . . . . ° f vibraThis equation 1S s1m1lar to the equatwn ° ·
tions of a body of mass m suspended on a spring
of rigidity k = a‘ot2/L:
ni:. = -—mg. ·—— kz.
· form harThis analo shows that the loop null per ,•
monic oscidihtions along the z·ax1s near the equi
Solutions 285
librium position determined by the condition
a*oc2 mgL
T Zo:—mg» Zo: *w.
The frequency of these oscillations will be
(
i/IH '
The coordinate of the loop in a certain time t
after the beginning of motion will be
_:& [ _ ( .22...
z ... dma 1-}-cos 1/% t .
3.55. The cross-sectional areas of the coils are
S1 = ::0%/4 and S2 = nD§/4. We shall use the
well-known formula for the magnetic flux (D =
LI = BSN, which gives B = LI/(SN). Therefore,
L 2 J1 is IN I ia
B2 L1 S2N2 I1 '
But I1 = I2 since the wire and Lthe current source
remain unchanged. The ratio of the numbers of
turns can be found from the formula N1/N2 =
D2/D1.- This gives
_Q;_= Lz5’iD¤ = L20;
B2 L1SgD1 LIDS °
Therefore, the magnetic induction in the new coil is
B = ———— ,
“ L,D2
3.56*. Let N1 be the number of turns of the coil of
inductance L1, and N2 be the number of turns of
the coil of inductance L2. It should be noted that
the required composite coil of inductance L can be
treated as a coil with N = N2 —l— N2 turns. If the
relation between the inductance and the number of
turns is known, L can be expressed in terms of L
and L2. For a given geometrical configuration of
the coil, such a relation must actually exist because
286 Aptitude Test Problems in Physics
inductance is determined only by geometrical configuration and the number of turns oi the coil (we
speak of long cylindrical coils with uniform wind1ng). Let us derive this relation.
From the superposition principle for a magnetic
field, it follows that the magnetic field produced
by a current I in a coil of a given size is proportional
to the number of turns in it. Indeed, t e doubling
of the number of turns in the coil can be treated as
a replacement of each turn by two new closely
located turns. These two turns will produce twice
as strong a field as that produced by a single turn
since the helds Hproduced by two turns are added.
Therefore, the eld in a coil with twice as many
turns is twice as strong. Thus, B GC N (B is the
maglnetic induction, and the current is fixed).
It s ould be noted that the magnetic flux embraced
by the turns of the coil is
¢D=BNS0€BN0¤N2,
It remains for us to consider that
L= -32- CX N 3.
Thus, we obtain L = kN“ for a given geometry.
Further, we take into account that N1 = 1/-L1/k,
N, = 1/L,/k, and hence L = k (N, -\- N,)*. Consequently,
L"—= VLILT
3.57. For a motor with a separate excitation, we
obtain the circuit shown in Fig. 224. In the first
case, i.e. when the winch is not loaded, 0 = I, =
(5 — $1)/r, where r is the internal resistance of
the motor, and $1 is the induced emf, 25, = owl.
Thus, ES, = 25, whence ot = 25/v1. In the second
case, the power consumed by the motor is
azz,.-.1......-..**5 "f’=”=· =mgv,.
Solutions 287
The induced emi is now $$2 = ow,. Thus, for the
internal resistance of the motor, we obtain
r __ (`5—ow2) oc
For the maximum liberated power, we can write
___$I I
where it can easily be shown that the maximum
power i5' = cw' is: liberated under the condition
6
Fig. 224
iS' = 8/2 (the maximum value of the denominator). Hence
v' = ai%:=·'éi·=2m’°·
·.....T£!........ .& N
m - 2(v1_v2) .-10 3 -6.7 kg.
3.58. Let us plot the time dependences UH (t)
of the external voltage, Ic (t) of the current in the
circuit (which passes only in one direction when
the diode is open), the voltage across the capacitor
U6 (t), and the voltage across the diode Ua (t)
(Fig. 225).
Therefore, the voltage between the anode and
the cathode varies between 0 and —2U0.
3.59. Since the current and the voltage vary in
phase, and the amplitude of current is I =
288 Aptitude Test Problems in Physics
UBX
UU " "* """° """"*’ ""‘°""-"'
0 AL AL
ir
._.. -, WV
`
Ic »
Un w6
0`
t
U" IIIII-I
YVY#
-U0 ..... ..... -.. - ....
““`“ "" '“``"`` """ '
I1
Fig. 225
250/R (i.e. the contributions from C and L are
compensated), we have
‘a}c·‘=°’L·
From the relations UC = q/C and dq/dt = I, we
obtain
_ $0 sin oat
”<> ·· "‘§a»F‘·
Solutions 289
Therefore, the amplitude of the voltage across the
capacitor plates is
QS mL
U° = "li2""·
3.60. During steady-state oscillations, the work
done by the external source of current must be
equal to the amount of»heat liberated in the resistor. For this the amplitudes ol the external
voltage and the voltage across the resistor must be
equal: RIO = UO. Since the current in the circuit.
and the charge on the capacitor are connected
through the equation I = dq/dt, the amplitudes
I0 and qc of current and charge can be obtained
from the formula
I0 _; w0q0·»
where the resonance frequency is mu =·= 1/
By hypothesis,
___ ___ Q1
U°‘ U ‘ T *
whence
Z zi E -. -. S3.61. For 0 < t < 1, charge oscillations will
occur in the circuit, and
q-(CU )cosm¢ 01-]/ 2
“’ "Tz" ° * °“’ 'Ec`*”·
At the instant ·r, the charge on the capacitor at its
breakdown is (CU/2) cos mor, and the energy of
the capacitor is (C U2/8) cosz wot. After the breakdown, this energy is converted into heat and lost
by the system. The remaining energy is
CU2 CU2
W ZJ •'*—··· ··•·· ··;····• ° ,
4 ( 8 )cos2w01:
19-0771 65
290 Aptitude Test Problems in Physics
The amplitude of charge oscillations after the
breakdown can be determined from the condition
W == q§/ (2C), whence
qc = —%·q- ]/2-—cos2 cont.
3.62. It is sufficient to shunt the superconducting
coil through a resistor with a low resistance which
can withstand a high temperature. Then the current
in the working state will pass through the coil
irrespective of the small value of the resistance
of the resistor. Il`, however, a part of the winding
loses its superconducting properties, i.e. if it has
a high resistance, the current will pass through
the shunt resistance. In this case, heat will be
liberated in the resistor.
4. Optics
4.1. Rays which are singly reflected from the
mirror surface of the cone propagate as if they were
emitted by an aggregate of virtual point sources
arranged on a c_ircle. Each such source is symmetrical to the source S about the corresponding
generator of the cone. The image of these sources
on a screen is a ring. It is essential that the beam
of rays incident on the lens from a virtual source
is plane: it does not pass through the entire surface
of the lens but intersects it along its diameter.
Therefore, the extent to which such a beam is absorbed by a diaphragm depends on the shape and
orientation of the latter. _
A symmetrical annular diaphragm (see Fig. 116)
absorbs the beams from all virtual sources to the
same extent. In this case, the illuminance of the
ring on the screen will decrease uniformly.
The diaphragm shown in Fig. 117 will completely
transmit the beams whose planes form angles
ot < ao with the vertical. Consequently, the 1l_luminance of the upper and lower parts of the ring
on the screen will remain unchanged. Other beams
will be cut by the diaphragm the more, the closer
the plane of a beam to the horizontal plane. _For
this reason, the illuminance of the lateral regions
of the ring will decrease as the angle cx varies between ot,) and ar/2. _
4.2. Let us first neglect. the size of the pupil, assuming that it is point-like. Obviously, only those of
the beams passing through the lens will get into
the eye which have passed through point B_before
they fall on the lens (Fig. 226). This point is conjugate to the point at which the pupil is located.
19*.
294 Aptitude Test Problems in Physics
to the geometrical path difference 6 due to the
deflection of the rays from the initial direction of
propagation. In this case. the interference of the
fw
•8\
•
y " r Ar
Fig. 228
rays will result in their augmentation (Huygens’
principle). It follows from Fig. 228 that
SOD, = in (r —{— Ar) - n (r)] l, 6 = Ar sin qi.
Hence
- _ 6 __ in-(r+Ar>—-Mr)! ___
sm q>--- -5-; -- ·----—-§——————— Z---- 2klr.
This leads to the following conclusion. If we consider a narrow beam of light such that the deflection
angle cp is small, then q> 0C r, i.e. the rotating vessel
will act as a diverging lens with a focal length
F = (2/lcl}'1.
Therefore, for the maximum deflection angle,
we obtain
Sin (Pmax == 2k¢ma.mConsequently, the required radius of the spot on
the screen is
R = rbeam + L ¢¤¤ emu-
Solutions 295
In the diverging lens approximation, we obtain
R ~ f1»eam+L<Pmax »’=“· f¤eam+2k!rneamL
2
="beam [i+°*PolL 14.4. The telescope considered in the problem is of
the Kepler type. The angular magnification k =
F/j, and hence the focal length of the eyepiece is
f == F/k == 2.5 cm. As the object being o served
approaches the observer from infinity to the smallest possible distance a, the image of the object
formed by the objective will be displaced from
the focal plane towards the eyepiece by a distance as
which can be determined from the formula for a
thin lens:
L+ I ___L 1 ____a——-F
a F—}—x — F ’ F—l—x — aF ’
aF , ~ F2
x"é`Z`-T ‘ ’ N T
since a > I•'. Thus, we must find zi:. The eyepiece of
the telescope is a magnifying glass. When an object
,._._-- ..;.....- ~.•- p
l -f. ,1...;Fig. 229
is viewed through a magnifier by the unstrained
eye (accommodated to inlinityl, the object must
be placed in the focal plane of the magnifier. The
required distance as is equal to the displacement of
the focal plane of the eyepiece during its adjustment. In this case, the eyepiece must obviously be
moved away from the objective. _ _
Figure 229 shows that when an inlimtely remote
object is viewed from the shifted eyepiece, the
298 Aptitude Test Problems in Physics
IJ1
Fig. 233
cylinder is sr/2, after refraction will form an angle
cz. with the cylinder axis such that sinu. = 1/n
(according to Snell’s law). The angle of incidence q>
of such a ray on the lateral surface of the cylinder
6`Meen
I
¤(
......l....-----.- - "
Fig. 234.
satisfies the condition oc -|- cp =-· at/2 (Fig. 234).
Since
. ri *1 y/5
“‘““*7·Ts·<—T·
oz < sr/4 and cp > an/4, i.e. the angle of incidence
on the lateral surface of the cylinder will be larger
than the critical angle of total internal reflection.
Therefore, this ray cannot emerge from the cylinder
at any point other than that lying on the right
base. Ani other ray emerging from the source
towards t e screen with a hole and undergoing refraction at the left base of the cylinder will propagate at a smaller angle [to the axis, and hence
will be incident on the lateral surface at an angle
exceeding the critical angle. Thus, the transparent
cylinder will "converge” to the hole the rays within
a solid angle of 2n sr.
In the absence of the cylinder, the luminous
flux confined in a solid angle of ndz/(4l)* gets into
the hole in the screen. Therefore, in the presence
of the transparent cylinder, the luminous flux
through the hole will increase by a factor of
2n
—————:[d,/(40, --8 >< 104.
4.9. The thickness of the objective lens can be found
from geometrical considerations (Fig. 235). Indeed,
2
r§=(2R1··-h) zi as 2R1h, ri: gg.,
where R1 is the radius of curvature of the objective.
0I
I
~" ‘ 4
{4% . F . `
M
V.
......"2_...
Fig. 235
Let us write the condition of ectiality of optical
paths ABF and CDF for the case w en the telescope
¢>n*
300 Aptitude Test Problems in Physics
is filled with water:
QU, -—— h) nw + 2hng; = nwll —|— h.
.Here fl is the focal length of the objective lens in
the presence of water. Substituting the values of h
and ll = ]/j§ -{— rj z fl -i— ri/(2f1), we obtain
22
gg; ~w=i,-gg [(¤g1—i)+(¤g1—-¤w)l·
Hence
f ___ R1”w
1 (”gl_1)+(”g1—nw) •
When the telescope does not contain water, the
focal length of the objective lens is
{gw = ....5;..... _
2 (**31*-1)
Therefore,
;,= fm _
° (¤g1—i)+(¤g1···¤w)
A similar calculation for the focal lengths fz
(with water) and {Q') (without water) of the eyepiece lens gives the following result:
js: {gv _
(ng1·· 1)-I-(¤g1—nw)
Therefore,
L=f1+f¤=(f{°’+fé°’) .
Since {Q2,) —l- ig';) = Lo, the required distance between the objective and the eyepiece is
... 2(ns1···i)”·w -
Solutions 301
4.10. Let the spider be at poirlt A (Fig. 236)
located above the upper point D of the sphere.
The spherical surface corresponding the arc BDB’
of the circle is visible to the sp der. Points B
lA
(6 Ai 5.
\
a` }'.-Q
a
fl
’
6 I c"
I
_,L
h" ’
I
M
7*.
Fig. 236
and B' are the points of intersection of tangents
drawn from point A to the surface of the sphere.
The ray AB"propagates within the sphere along BC.
The angle oc can be found from the condition
.1
SID <Z= —-—— ,
where ng] is the refractive index of glass. This ray
will emerge from the sphere along CA' Therefore,
the fraction of the spherical surface corresponding
to the arc CD’C' will also be visible (by way of
302 Aptitude Test Problems in Physics
an example, the optical path of the ray AKLM
is shown).
The surface of the spherical zone corresp.onding
to the arcs BC and B'C’ will he invisible to the
spider.
The angle 9 is determined from the condition
S ·-...-.'E.-where R is the radius of the sphere, and h is the
altitude of the spider above the spherical surface.
Since R > h by hypothesis, ·y z 0. We note now
that B = n — 2a and sin on = 1/ng]. Therefore,
.1n
B...rc—2arcs1n(7-I-5-) N -2-,
Thus the ogposite half of the spherical surface is
visible to t e spider, and the fly must be there.
4.11. None of the rays will emerge from the lateral
surface of thegcylinder if for a ray with an angle of
-_]L.¤. .
.,5
S 7* .
I
Fig. 237
incidence y z at/2 (Fig. 237), the angle of incidence
on on the inner surface will satisfy the relation
sin oc >1/n. In this case, the ray will undergo
total internal reflection on the lateral surface.
It follows from geometrical considerations that
sin or.= }/1——sin* B, sin 5:%- ,
Solutions 303
Thus,
nmin: l/2
4.i2. By hypothesis, the foci of the two lenses are
made to C0iI1Cid€, i.0-. UIC S€p8l'3t·i0I1 b€l3W08I1 thé
lenses is 3f, where f is the focal length of a lens with
a lower focal power.
In the former case, all the rays entering the tube
will emerge from it and form a circular spot of
radius r/2, where r is the radius of the tube
(Fig. 238). Inlthe latter case, only the rays which
Pv
f2
z <al·
4?‘
Fig. 238
enter the tube at a distance smaller than r/2 from
the tube axis will emerge from the tube. Such
rays will form a circular spot of radius r on the
screen (Fig. 239). Thus, if J is the luminous in2é
7I
z> é 4
?(;
Fig. 239
tcnsity of the light entering the tube, the ratio
of the illuminances of the spots before and after
306 Aptitude Test Problems in Physics
. lz —-— hz nz sinz (on/2)
Therefore, 1f ——-§——— > 1 _ nz Sinz (OL/2) , the
required time is
t—2 ( v + v/n )
.. EE. (..................‘
U t/1-—— nz sinz (oc/2)
+ n t/lz—hz sin (oc/2) _ nz sinz (oz/2) )
h t/1 ——nz sinz (oc/2)
___2h V .“"6i` i/z=-n= . n
lz —— hz nz sinz (ot/2) __ 2l
H hz < 1 ·— nz sinz (oz/2) ’ thm t`--7 _
4.15. It follows from symmetry considerations
that the image of the point source S will also be
S s' C
,...0.... -...Tm..- .. ... ... .-•-.-.
‘·\` .
Fig. 241
at a distance b from the sphere, but on the opposite
side (Fig. 241).
4.16. An observer on the ship can see only the rays
for which sin on <1/n 1 (if sin on > 1/n 1, such
a ray undergoes total infernal reflection and cannot
be seen by the observer, Fig. 242). For the angle B,
we have the relation
nw sin B--: ng; sin rx, sin 5:-2-Q- sin oz,
nw
Solutions 307
where ng] is the refractive index of glass. Since
I sin cz. I < 1/ngl, I sin B I < 1/nw. Therefore, the
observer can see only the objects emitting light
J ‘—" 1 i—*;—;—‘
_· i Z `—
Fig. 242
to the porthole at an angle of incidence B Q
arcsin (1/nw). Figure 242 shows that the radius
of a circle at the sea bottom which is accessible to
observation is R z h tan B, and the sought area
(li tan B> D/2) is
z
S=:tR2z éq-I-·—— z82m’.
nw-—-1
4.17*. Short-sighted dpersons use concave (diverging) glasses which re uce the focal power of their
eyes, while long·sighted persons use convex (converging) glasses. It is clear that behind a diverging
lens, the eye will look smaller, and behind a converging lens larger. If, however, you have never
seen your companion without glasses, it is very
difficult to say whether his eyes are magnified or
reduced, especially if the glasses are not very strong.
The easiest way is to determine the displacement
of the visible contour of the face behind the glasses
relative to other parts of the face: if it is displaced
inwards, the lenses are diverging, and your companion is short-sighted, if it is displaced outwards,
the lenses are converging, and the person is longsighted.
304 Aptitude Test Problems in Physics
the reversal of the tube is
J J /4 E, 1
E1" n(r/Z)2 ’ E2 mz ’ E1 " 16 °
4.13. The light entering the camera is reflected
from the surface of the a ade. It can be assumed
that the reflection of ligilxt from the plaster is
practically independent of the angle of reflection.
In this case, the luminous energy incident on the
objective of the camera is proportional to the solid
angle at which the facade is seen from the objective.
As the distance from the object is reduced by half,
the solid angle increases by a factor of four, and
a luminous energy four times stronger than in the
former case is incident on the objective of the same
area.
For such large distances from the object, the
distance between the objective and the film in the
camera does not practically change during the
focussing of the object and is equal to the focal
length of the objective. The solid angle within
which the energy from the objective is incident
on the surface of the ima e depends linearly on
the solid angle at which Sie facade is seen, i.e.
on the distance from the object. In this case, the
illuminance of the surface of _the image (which,
by hypothesis, is uniformly distributed over the
area of this surface), which determines the exposure, is di1·ect1y proportional to the corresponding energy incident on the objective from the facade
and inversely proportional to the area of the image.
Since this ratio is practically independent of the
distance from the object under given conditions
there is no need to change the exposure.
4.14*. The problem is analogous to the optical
problem in which the refraction of a plane wave
in a prism is analyzed. According to the laws of
geometrical optics, the light ray propagating from
point A to point B (Fig. 240) takes the shortest
time in comparison with all other paths.
The fisherman must move along the path of a
"1ight ray", i.e. must approach point E of the bay
at an angle y, cross the bay in a boat at right angles
Solutions 305
to the bisector of the angle ot, and then move along
the shore in the direction of point B.
.----...
/
t
QT \.,’/ ® ,» ’“/Z
¤ .. . B
/
\\ C
" * sl
..\
Fig. 240
The angle Y can be determined from Snell’s law
(n = Z):
sin Y — n sin OL
" T`
The distance a is
azhmnvzh nsin(oc/2)
]/1-nz sinz (oc/2) O
The distance b can be determined from the equation a —}— b = ]/lz — hz. Hence
b-:. 1/l2...-h2....h
]/1-nz sinz (oc/2) •
If b 0 _ 9 2 2 nz sinz (ot/2)
> y l•€• Z "' h > h 7
the fisherman must use the boat. Separate segments
in this case will he
EK = p ··= b sin —%—
=_-( "/[2._h2.s..h ) Sin .2..,
_ }/1-nz sinz (oz/2) 2
AE = q =. ....*3..../..= ........L*.,....... _
°°S Y V 1 —n* s1n* (ct/2)
308 Apt.itude Test Problems in Physics
4.18. We decompose the velocity vector v of the
person into two components, one parallel to the
mirror, v,,,f_and the other perpendicular to the
/
/
/¤
I
’,
Un [ i U Uri ] U"
"r ?<· ’°.L
Fig. 243
mirror, vl, i.e. v = vu —|— vl (Fig. 243). The
velocity of the image will obviously be v' =··
vn —— vJ_. Therefore, the velocity at which the
person approaches his image is defined as his
velocity relative to the image from the formula
vm] = 2v_L = 2v sin ot.
4.19. Let O be the centre of the spherical surface
of the mirror, ABC the ray incident at a distance
BE from the mirror axis, and OB = R (Fig. 244).
From the right triangle OBE, we find that sin oc ==
h/R. The triangle OBC is isosceles since LABO =
LOBC according to the law of reflection, and
4BOC = LABO as alternate-interior angles.
Hence OD = DB = R/2. From the triangle ODC,
we obtain
,,, .. ....’L.. - ...E._...
— 2cosot — 2 1/R2..h·.:
(C is the point of intersection of the ray reflected
by the mirror and the optical axis).
For a ray propagating at a distance hl, the distance :1 z R/2, with an error of about 0.5% since
Solutions 309
h%< R2. For a ray propagating at a distance hz,
the distance x2 = 3.125 cm. Finally, we obtain
A:c=x2—x1zO.6cm#O(!)
A _ ,6
ez
2 0,
• 0 d. \\ .
x .x
Fig. 244
4.20. Let us consider a certain luminous. point A
of the hlament and an arbitrary ray AB emerging
°x
x
//
x
»x
//
x
xW
"""'4(-—'°—_°—',,—. / /
// [zzz 0 \\ 0
//z·’· \
AI
Fig. 245
from it. We draw a plane through the ra y and the
filament. It follows from geometrical considerations
310 Aptitude Test Problems in Physics
that with all possible reflections, the given ray
will remain in the constructed plane (Fig. 245).
After the first reflection at the conical surface,
the ray AB will propagate as if it emerged from
point A', viz. the virtual image of point A. The
necessary condition so that none of the rays emerging from A ever gets on the mirror is that point A'
must not be higher than the straight line OC, viz.
the second generator of the cone, lying in the plane
of the ray (point O is the vertex of the conical surface). This will be observed if
.LA’0D—}—L.AOD—l—LAOC-=3 -%- ;· i80°.
Consequently,
amln > 1200-
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